Let me analyse this for you: You anchor yourself to the station. So your weight is now in the ballpark of $m_1=420\times 10^3\text{kg}$. The shot bullets are about $m_2=8\times 10^{-3}\text{kg}$ each and travel at $v_2=710 \frac {\text m} {\text s}$.
Insert these into these:
- $F_1=5\times F_2=m_1\times a_1 = m_1\times\frac {dv_1}{dt_1}$
- $F_2=m_2\frac {dv_2}{dt_2}$
Assume $dt_2=0.1\text s$, $dt_1=5\times dt_2$ and do some simple math that I won't give here and you get...
- $dv_1=\frac{5\frac{8\times10^{-3}}{420\times10^3}\frac{710 \frac {\text m} {\text s}}{0.1\text s}}{5 \times 0.1\text s}=0.00135\frac {m}{s}$
Up to here, only the numeric value of the change in speed was relevant, NOT the direction (vector) at all. To see how this affects the rotation, you'll have to insert some data: your own position from that rotational axis and the vectors of both the shot and the station.
- $\vec {v_1}=\vec\omega_1\times\vec{r_2}$
- $\vec{d\omega_1}=\frac{\vec {dv_1}} {|\vec{r_2}|}$
This change in orbit (which is dependant on the part of $\vec{v_1}$ along its orbit only) and rotation ($\vec {d\omega}$) is tiny and the next automated correction burn will fix more than this.