On sufficient conditions to ensure $X/G\cong X/G'\Rightarrow G\cong G'$.

Let $X$ be a topological space and let $G$ and $G'$ be two groups acting on $X$ such that $X/G\cong X/G'$.

Question. Under what conditions can we conlude that $G$ is isomorphic to $G'$?

Counter-example. Two non-isomorphic groups acting trivially on $X$.

Sufficient condition. $X$ simply-connected and $G$ and $G'$ acting freely and properly discontinuously on $X$. Under these assumptions, $G$ and $G'$ are isomorphic to the fundamental groups of $X/G$ and $X/G'$.

One of my goal is to drop out the assumption on the simple-connectedness of $X$, but then the remaining assumptions are not strong enough as it shown with $X=\mathbb{S}^1$, $G=\{1\}$ and $G'=\mathbb{Z}/2\mathbb{Z}$.

Any help will be greatly appreciated.

In general, this is quite hopeless. For instance, suppose that $X$ is any closed connected oriented surface whose Euler characteristic equals $-2pq$, with $p\ge q>1$. Then $X$ always admits covering actions $G\times X\to X, G'\times X\to X$ of nonisomorphic abelian groups (of the orders $=pq$), with $X/G\cong X/G'$, the genus 2 surface.

Things get a bit better if you assume that $G, G'$ are finite subgroups of $Homeo(X)$ acting as groups of covering transformations, so that $G\le G'$ (like in your example with the circle). Suppose that $c(Y)$ is a numerical invariant of (some class of topological spaces) which is multiplicative under covering maps: if $p: Y_1\to Y_2$ is a covering map of degree $d$, then $c(Y_1)=d c(Y_2)$, where $d$ is the degree of $p$ (i.e. cardinality of point-preimages $p^{-1}(y), y\in Y_2$). For instance, $c$ could be some characteristic number, e.g. the Euler characteristic or a Pontryagin number. Or maybe $c$ is the Gromov-norm. Assume also that $c(X)$ exists, is finite and nonzero. Then in our setting, $$ c(X)= |G| c(X/G) = |G'| c(X/G') $$
and the existence of a homeomorphism $X/G\to X/G'$ implies that $|G|=|G'|$. Since $G\le G'$, it follows that $G=G'$.