The angular resolution of the naked human eye is approximately one arc minute, or 1/60th of a degree. (Also a nice illustration of what this means in practice, which just so it happens also shows the ISS for scale.) This is going to be different for different people, but should serve as a baseline (and it is also a convenient number to work with).
Given an angular size $a$ (in seconds of arc), object diameter $d$ and distance $D$, we know that $$ \frac{a}{206265} = \frac{d}{D} $$ where 206,265 is the number of arc seconds in one radian ($ 206265 = 60 \times 60 \times \frac{360°}{2\pi} $) and $d$ and $D$ are in the same units. We know that $ a = 60 $ (arc seconds) and $ D = 400 \text{ km} $, and if we plug these in and use the here more useful unit of meter then we get $$ \frac{60}{206265} = \frac{d}{400000 \text{ m}} $$
Solving for $d$ gives us $$ d = 400000 \text{ m} \times \frac{60}{206265} \approx 116 \text{ m} $$ which tells us that the smallest resolvable size for the human eye at distance 400 km is approximately 116 m. This is, in a manner of speaking, the "pixel size" of the human eye at that particular distance.
Hence, any spacecraft features must be at least 100 meters or so large for them to be discernable by the naked eye at a distance of 400 km. For comparison, the International Space Station measures 109 meters truss length by 73 meters solar array length and nominally orbits at 418 to 423 km altitude. While the ISS is visible, its shape is not discernable.
Note that the size figures you get for this assumes that $D$ is the distance to the object. If the spacecraft is directly above the ground-based observer, the distance will be equal to the altitude, but if the spacecraft is not directly overhead then the distance to it from the observer will be larger. (Trigonometry is your friend here.)
Also note that the brightness of the object makes a potentially very large difference. Many objects smaller than this theoretical value are visible to the naked eye because they put out, or reflect, enough light to be registered by the rods or cones of the eye; however, those appear as point light sources without any discernable features, just like remote stars. A spacecraft is likely to be made of reflective material, which will be reflecting sunlight making it much easier to spot.
Atmospheric conditions, light pollution and other factors will also contribute to whether an object is visible. Clear wilderness skies are obviously better for this purpose than light- and smog-polluted city landscapes, for instance.
Any individually discernable features will need to be at least this minimum size however to be clearly visible. Note that sunlight reflecting off the spacecraft may make individual features more difficult to make out even if they nominally might be large enough, so you may need to make them even larger to be readily visible.
A spacecraft in low Earth orbit will move very differently compared to a star, so assuming that it is visible in the first place, it will be easy to tell that it is not a star. Primarily, it will be moving much faster across the observer's field of vision than any star would.
For the grand finale, borrowing from TildalWave's answer on the ISS over on Space Exploration also linked to earlier:
According to Human Photoreceptor Topography, Curcio et al., 1990 (PDF) that lists several sources as well as own measurements of the spatial density of cones and rods in whole-mounted human retinas, the greatest cone density recorded was 324,100 cones/mm2. That gives us acuity (or row-to-row spacing we need to discern at least two individual features) of 86.3 cycles/°. So for our best case, with a great eye, neglecting any atmospheric effects, the ISS right above us when it's closest, and optimal contrast with the background sky, we get minimum separation of objects of 74.83 m. If there was no air between the observer and the station!
Do note that TildalWave uses 370 km as the ISS' orbital altitude, which is less (though not significantly less) than the 400 km used in this question. Said differently, the two are in close enough agreement for our purposes: at 400 km distance, ignoring such pesky details as air, you need features to be at least about 100 meters in size in order to be visible to the naked eye (and that is an absolute minimum). Allowing for the obscuring effects of air, and quite possibly for the problem of making things out near a bright point light source such as the surface of a spacecraft in sunlight, you need to make them larger than that.
