Let $X$ and $Y$ be locally compact topological spaces, both equipped with a proper $G$-action where $G$ is a locally compact group. Let $f: X \to Y$ be a $G$-equivariant local homeomorphism.
Is the quotient map $\bar{f} : X/G \to Y/G$ still a local homeomorphism?
I think that it is not hard to see that $\bar{f}$ is continuous and open, but I guess it does not have to be locally injective, though I can't think of a counter-example.
This is false already for proper actions of discrete groups; the point is that the action $G\times X\to X$ can be free while the action $G\times Y\to Y$ non-free. As the result, the map $\bar f$ can fail to be a local homeomorphism.
As an example, consider $G={\mathbb Z}/2$ acting on ${\mathbb R}$ so that the generator acts as $x\mapsto -x$; $X$ is the disjoint union of two intervals $[-3,-1]\sqcup [1,3]$, while $Y=[-1, 1]$ with the same $G$-action as above. The map $f$ restricts to the translation by $\pm 2$ on each of the components of $X$ and, hence, is a homeomorphism on each of the components. The map $\bar{f}$ can be identified with the absolute value map $h: [-1,1]\to [0,1]$.