Answer
The equation is only symmetric with respect to the line $\theta=\frac{\pi}{2}$.
see graph.
Work Step by Step
Step 1. We are given the polar equation $r=-4sin\theta$. To test the symmetry with respect to the polar axis, let $\theta\to -\theta$; we have $r=-4sin(-\theta)=4sin\theta$. Thus the equation is not necessarily symmetric with respect to the polar axis.
Step 2. To test the symmetry with respect to the line $\theta=\frac{\pi}{2}$, let $r\to -r$ and $\theta\to -\theta$; we have $-r=-4sin(-\theta)$ or $r=-4sin\theta$. Thus, the equation is symmetric with respect to the line $\theta=\frac{\pi}{2}$.
Step 3. To test the symmetry with respect to the pole, let $r\to -r$; we have $-r=-4sin\theta$ or $r=4sin\theta$. Thus, the equation is not necessarily symmetric with respect to the pole.
Step 4. We can graph the equation as shown in the figure.
