Let $G$ acts on space $X$, We say the action of $G$ on $X$ is proper discontinuous, if for every $x\in X$, there is an open set $U$ containing $x$ such that $gU\cap U=\phi$ for all $g\in G, g\neq e$
We say the action is $\textit{covering space action}$ if $G$ acts by homeomorphism and action is proper discontinuous.
Now Let $M$ be a metric space, Let $G$ be a group which acts on $M$ by isometry.
SO my question is the following, What extra conditions are required to conclude that this action is a covering space action.
$\textbf{My attempt}:$
Since, Isometry is a homeomorphism, we conclude that $G$ acts on $M$ by homeomorphism. We only need to make sure that the action is proper discontinuous.
Now, If we add a extra condition, that action is fixed point free, ($g.x=x \implies g=e$), then I guess we can conclude that action is covering space action.
Proof: Assume that action is not p.d.c., Let $y\in M$, and $U=B(y,\frac{1}{n})$ be an open set containing $y$. Since the action is not p.d.c there exist $x\in gU\cap U$.
$d(x,y)<\frac{1}{n}$, also $x\in gU$, so $x=g.z$ for some $z\in U$.
$d(z,y)<\frac{1}{n}$,
$d(g.z,z)\leq d(g.z,y)+d(z,y)=\frac{2}{n}$. We can choose $n$ large enough. Hence we arrive at a contradiction as $z$ is a fixed point.
Do I need to put another condition, that orbits need to be discrete. What If I don't assume it. Is there a counterexample where action of $G$ on $M$ is by homeomorphism and action is fixed point free, but still it is not a covering space action.