No.
To my knowledge, the only really serious calculations regarding this scenario are in an article by Alexander Bolonkin and Joseph Friedlander. It's currently cited in the current highest-voted answer as a feasibility study of the possibility of destroying the Sun by detonating a nuclear weapon in the Sun's atmosphere, inducing a self-supporting nuclear detonation wave that would subsequently propagate throughout the entire Sun, causing a catastrophic explosion. I think it's an excellent guide with which to show that this idea is not at all possible, contrary to the claims made. Given that, I'm going to critique its analysis, and therefore the scenario given.
The setup
Let's assume that someone has created a spaceship, placed a nuclear weapon on board, and sent it on a trajectory towards the Sun. They've timed it to detonate in the solar atmosphere; moreover, they've designed shielding that protect it from high temperatures and solar activity like flares and coronal mass ejections. Essentially, we can assume that the payload is delivered successfully and the detonation begins as desired.
If a nuclear weapon was detonated in any environment, creating a self-sustaining blast wave, the wave would be supported by whatever fusion reactions are favored by the surrounding medium. In other words, the weapon itself doesn't dictate the type of nuclear reactions supporting the blast wave, and the most efficient ones will be chosen. This is something that was studied during the Manhattan project. The scientists were concerned that the first detonation of a nuclear weapon would initiate a self-supporting blast wave that would travel through the atmosphere and oceans, killing all life on the planet.
It's a scary possibility, and naturally, it was modeled in a lot of detail. A number of papers were published on it over the years, including Ignition of the atmosphere with nuclear bombs. In air, the reactions the physicists were most concerned about involved the fusion of two nitrogen atoms - certainly a possibility, as nitrogen is the most abundant component of the atmosphere. Even though the groups considered the most favorable conditions for sustaining such a blast wave, they found a runaway detonation impossible for reasonably powerful nuclear weapons. I'm sure they thoroughly checked their calculations.
The Sun is largely composed of hydrogen, ionized because of the high temperatures. It generates energy primarily via a form of the proton-proton chain reaction (p-p chain); much higher temperatures would be needed to use reactions found in more massive stars. In particular, a variant called the p-p I branch is dominant and most temperatures in the solar core. It's reasonable to expect that the same sort of reactions would occur immediately following the detonation of the weapon, provided the required temperatures (10-15 million Kelvin) could be achieved.
Why would a nuclear weapon help?
With the exception of the corona, the Sun's photosphere has a temperature of about 5800 K. The temperature increases further into the Sun, but with the exception of the core, conditions aren't extreme enough for nuclear fusion to proceed. Bolonkin claims that even in the core, temperatures are low enough that the p-p chain proceeds slowly - about 15 million Kelvin. He invokes something called the Coulomb barrier to support his point, claiming that a nuclear weapon could surmount it.
The Coulomb barrier is an extremely well-studied phenomenon, because it's extremely important when fusion is on the verge of happening. Nuclei have a net positive charge, as they're composed of protons. Therefore, any two nuclei will repel each other if brought close together, via the electrostatic force - described by Coulomb's law, which you've probably talked about in an introductory physics course. This repulsion gets stronger the closer together the nuclei get, meaning that it's very, very hard to overcome the force. This is the Coulomb barrier.
The Coulomb barrier is a problem - so big a problem, in fact, that stars shouldn't be able to avoid it. Stellar fusion would be impossible except at extremely high temperatures - over 10 billion Kelvin! Fortunately, there's a way around it: quantum tunneling. Quantum tunneling arises because a particle's position and momentum can never be known exactly, and there is always a probability that a particle will be in a given location. The wavefunction of the particle - a description of how likely it is to be in a certain state - shows that two protons have a probability of being arbitrarily close together, which would normally be forbidden by classical physics.
Bolonkin ignores quantum tunneling, arguing that the merit of a nuclear weapon is that it could temporarily raise temperatures in a small region of the Sun. The higher the temperature, the more likely a particle is to move at higher speeds. Therefore, more protons would be likely to fuse. I've seen the same logic used elsewhere to justify using a nuclear weapon in this scenario. However, the temperatures around a nuclear weapon will only rise to several tens of millions of Kelvin - extremely hot by most standards, but far too cool to help more particles overcome the Coulomb barrier.
The stability conditions
Bolonkin claims that in order for a detonation wave to continue propagating, it must travel faster than the ion speed of sound. He eventually derives what he claims is the criterion for a successful, self-sustaining blast wave:1
$$n\tau>\frac{\gamma zk_BT}{(\gamma^2-1)E\langle\sigma v\rangle}\tag{1}$$
where:
- $n$ is the number density of particles.
- $\tau$ is something equivalent to the confinement time
- $\gamma$ is the adiabatic index
- $k_B$ is the Stefan-Boltzmann constant
- $T$ is the temperature of the environment
- $E$ is the energy of the reaction
- $\langle\sigma v\rangle$ is the mean reaction rate - an average of the product of the collisional cross-section of a proton and the relative velocity of protons
- $z$ is the charge of the nucleus divided by the fundamental charge.
Bolonkin claims that his condition is superior to the Lawson criterion, which is commonly used in designs of nuclear fusion reactors to determine whether fusion can take place. It's usually derived from a perspective of energy loss: Can the reaction, in the given environment, produce more energy than it loses? The Lawson criterion is
$$n\tau>\frac{12k_BT}{E\langle\sigma v\rangle}\tag{2}$$
which is very similar. The authors seem to imply that Lawson's derivation is inapplicable in a star because, as they claim, there are no energy losses; in a nuclear reactor, on the other hand, energy can be lost to the walls and surrounding environment. Therefore, they conclude, their version is correct. Well, then let's see how much more favorable their condition is. Bolonkin says that $\gamma$ should be between 1.2 and 1.4, and that $z$ should be set to 1. In the cases where $\gamma=1.2$ and $\gamma=1.4$, we find that
$$n\tau>\frac{2.73k_BT}{E\langle\sigma v\rangle},\quad n\tau>\frac{1.46k_BT}{E\langle\sigma v\rangle}$$
That's not a huge improvement - lower than Lawson's by a factor of 4 to 8, roughly. We shouldn't get too excited here. It's debatable as to whether either criterion holds, in fact, as Bolonkin failed to consider energy losses in the photosphere, where the detonation would originate. The upper layers of the Sun's atmosphere are optically thin, meaning that light can travel through them with relative ease. I'm concerned reasonable that, therefore, energy would be lost rather easily. Slightly more complex formulations of the Lawson criterion look at other sources of energy loss; Bolonkin's clearly does not.
One form of energy loss that comes to mind is thermal bremsstrahlung. Bremsstrahlung is radiation emitted when one charged particle is accelerated or decelerated by another. Given that after the detonation, we have hot ($\sim10^7$ Kelvin) plasma in an environment that may be optically thin to these x-rays, bremsstrahlung could be an efficient form of energy loss.2
I should note, of course, that the Lawson criterion is usually applied to nuclear reactors, not stars. Therefore, it seems strange that Bolonkin would want to compare his results to Lawson's at all.
The thermostat effect
The Sun is composed mostly of plasma - largely, as I said above, of hydrogen nuclei - protons! The gas obeys the ideal gas law, hopefully another concept you've come across before. The ideal gas law is an equation of state, meaning that it relates several thermodynamic variables together. Although the law is usually formulated as $PV=nRT$, a sometimes-preferred form in astrophysics is
$$P=nk_BT\tag{3}$$
where $P$ is pressure, $n$ is number density, and $T$ is temperature. The ideal gas law should hold well in the outer layers, and should be a decent approximation in the core. The big criterion is that the thermal energy be much larger than the energy of interactions between protons, which holds in general. The standard solar model confirms this; the ideal gas law's predictions largely agree.
There are some pretty nice consequences of the ideal gas law. Let's say that temperature in a pocket of the Sun rises, thanks to the rate of nuclear reactions increasing. This should in turn speed up the reaction rate; I said before that higher temperatures are more beneficial to fusion. Well, according to the ideal gas law, if the temperature rises, then either the pressure increases or the density decreases.
It turns out that we should expect $P$ to increase and $n$ to decrease simultaneously. A star supporting itself by nuclear fusion is in hydrostatic equilibrium. The gas pressure trying to expand the star opposes the force of gravity trying to collapse the star. However, if the temperature rises, the pressure will increase. Suddenly, the star is out of equilibrium, and the net force on any layer will be upwards, away from the center. This lowers the density, which in turn lowers the reaction rate and the temperature, bringing the star to equilibrium again. This is sometimes informally referred to as the solar thermostat. This prevents runaway nuclear reactions, for the most part.
The quantity $\langle\sigma v\rangle$ is often approximated as being a power law in terms of temperature dependence. That is, $\langle\sigma v\rangle\propto T^\eta$, where $\eta$ is a constant. For the p-p chain, there is a small temperature dependence, relative to other reactions (like the CNO cycle). In particular, we can say that $\eta=4$.3 If we plug this into either version of the criterion, we find that
$$n\tau>CT^{-3}$$
where $C$ is a constant depending on which criterion you've chosen. Therefore, at lower temperatures, $n\tau$ must be greater, making it harder and harder for fusion to occur as the temperature drops. Again, this assumes that both criteria are valid; even if they are, the risk of a runaway detonation is non-existent.
Astronomical events
It turns out we can look to the skies to think about naturally-occurring events that are similar to the scenario you describe. First, there are examples of solar activity, including solar flares and coronal mass ejections. The energy released in these events can range from $\sim10^{20}$ Joules to $\sim10^{25}$ Joules. However, the Tsar Bomba (the most powerful nuclear weapon ever detonated) released only $\sim10^{17}$ Joules. Given that solar flares regularly release thousands of times as much energy in the photosphere - the target region of detonation - without any catastrophic problems, I think we can consider the risk of detonation by nuclear weapon to be even lower.
Moving on, consider helium flashes. These are believe to occur in low-mass red giants (less than 2 solar masses). As hydrogen fusion ceases in the core of a star (while continuing further out), the core falls out of hydrostatic equilibrium, and the star begins to contract. This raises temperatures until matter in the core becomes degenerate. Degenerate matter does not obey the ideal gas law,4 and so cannot fight back against rising temperatures. Eventually, runaway fusion begins via the triple alpha process, at temperatures around 100 million Kelvin. However, even under such conditions, the matter soon becomes non-degenerate. Thermal pressure returns, the ideal gas law is applied, and the star is in hydrostatic equilibrium once more. Helium flashes are much more powerful than solar flares, coming in at around $\sim10^{41}$ Joules. You can read more about the instabilities involved in these detailed slides.
The thermostat mechanism is not applicable in objects composed solely of degenerate matter, like white dwarfs. This often has dire consequences; if matter is transferred onto the surface of a white dwarf and it heats up, runaway fusion can occur, usually involving carbon and oxygen. The result is a nova, which leaves much of the star intact, or a Type Ia supernova, which may destroy the white dwarf or turn it into a neutron star or black hole. Type Ia supernovae usually release $\sim10^{44}$ Joules of energy - although this is a byproduct of a successful detonation, not the cause of it.
Numerical simulations have been done of the propagation of detonation waves through white dwarfs. One result is that detonations have the potential to turn into deflagration waves, which are less catastrophic. This has been studied a lot in pure fluid dynamics, but it's interesting to know that instabilities can quench possible detonations in white dwarfs - I'll try to pull up an article on some examples. It makes me wonder whether, even if I'm wrong about everything above, if this hypothetical detonation could falter into a deflagration, therefore saving the Sun from destruction.
However, even in extremely catastrophic situations, a non-degenerate star like the Sun can stabilize itself against runaway fusion reactions. A red giant could survive a helium flash, which at first seems extremely devastating. There's no way that a puny nuclear weapon could overcome the mighty thermostat effect. In short, if you're trying to blow up the Sun, I'd recommend turning your efforts elsewhere. Bolonkin and Friedlander are, simply put, wrong.
Footnotes
1 His notation is non-standard and unclear, and include unnecessary terms for unit conversions. I've standardized them here for clarity, and fixed a typo or two he made.
2 The power radiated by thermal bremsstrahlung is proportional to $T^{1/2}$.
3 We call the case where $\eta=4$ weakly temperature dependent because some fusion reactions in slightly more massive stars involve $\eta=17$ or $\eta=20$!
4 White dwarfs and matter supported by electron degeneracy obey one of two main equations of state. For the ideal gas law, $P\propto\rho T$, where $\rho$ is density. White dwarfs obey $P\propto\rho^{5/3}$ (non-relativistic) or $P\propto\rho^{4/3}$ (relativistic), depending on the regime. In both cases, there is no temperature dependence.