Let $X$ be a metrizable topological space, and suppose we have three groups $\Gamma$, $\Gamma'$, and $\Gamma''$, acting properly on, respectively $X$, $X$, and $X/\Gamma'$. What can we say about them if we know that $X/\Gamma\cong (X/\Gamma')/\Gamma''$?
My intuition is that the latter two contains together all the information of the first, so I would be tempted to say that we have a short exact sequence $0\to\Gamma'\to\Gamma\to\Gamma''\to 0$. In fact this would be the analogue of the third isomorphism theorem.
This setting appears when studying orbifold coverings, so we might assume all groups being finite subgroups of $O(n)$, and $X$ an open subset of $\mathbb{R}^n\times\mathbb{R}^{\geq 0}$, but I suspect the answer could hold much more generally.
Take $\Gamma'' = 1$ for simplicity. Then you are asking whether $X/\Gamma \cong X/\Gamma'$ implies $\Gamma \cong \Gamma'$. Even if we assume that the actions of $\Gamma$ and $\Gamma'$ are free as I suggested in the comments this need not be true in general; for example, any cyclic group $C_n$ acts freely by rotations on $X = S^1$ with quotient $S^1$ again.
However, it is true if $X$ is simply connected and we assume the actions of $\Gamma$ and $\Gamma'$ are properly discontinuous, since then $\Gamma \cong \Gamma'$ is the fundamental group. This gives us a salvage of the general case $\Gamma'' \neq 1$; if we assume that $X$ is simply connected and all three actions are properly discontinuous then the fundamental group of $(X/\Gamma')/\Gamma''$ sits in a short exact sequence $1 \to \Gamma' \to \pi_1 \to \Gamma'' \to 1$ (e.g. by the long exact sequence in homotopy) so we have a short exact sequence $1 \to \Gamma' \to \Gamma \to \Gamma'' \to 1$ as desired.