does isomorphic groups induce homeomorphic quotients suppose $X$ is a topological space and $G$ and $H$ are groups acting on it.
1) if $G$ is isomorphic to $H$ do we have necessarely $X/G$ is homeomorphic to $X/H$
2) suppose $G$ and $H$ are two conjugate subgroups of a group $K$ acting on $X$, in this case do we have $X/G$ is homeomorphic to $X/H$
3) Conversely, if $X/G$ is homeomorphic to $X/H$ does this implie that $G$ is isomorphic to $H$?
 A: The following are hints:
(1) Consider the case $G=H$ and consider one action of $G$ on $X$ to be defined by $g\cdot x=x$ for all $x\in X$. (Prove that this is indeed an action of $G$ on $X$ if you have not done so already.) The quotient space $X/G$ is naturally homeomorphic to $X$. Find an example of a non-trivial topological group $X$ and an action of $G$ on $X$ such that $X/G$ has exactly one orbit (for example). Conclude that $X/G$ is not homeomorphic to $X/H$.
(2) Let $k\in K$ be such that $k^{-1}Gk=H$. Prove that $x\mapsto k\cdot x$ induces a homeomorphism $X/G\to X/H$. (In particular, you need to prove that if two elements of $X$ lie in the same $G$-orbit, then their images under this map lie in the same $H$-orbit. For this, use $k^{-1}Gk=H$.) 
(3) Consider two non-isomorphic groups acting trivially on $X$.  
A: First a question. When you talk about `action of a group on a topological space' is it assumed that the action is continuous?


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*No. You could  let $\mathbf{Z}$ act on $\mathbf{R}$ by translation. Then $\mathbf{R}/\mathbf{Z}$ is homeo to the circle. If we let $\mathbf{Z}$ act on $\mathbf{R}$ trivially, we just get $\mathbf{R}/\mathbf{Z} = \mathbf{R}$. 


So the point is that the same group can give many different actions and, therefore, many different quotients.


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*No. (I'm not sure if the following example works.) Let G be the cyclic group of order two acting on the circle S^1 via multiplication with -1.  Let H be the trivial group acting trivially on the circle. Note that S^1/H = S^1 and S^1/G$ is (also?) homeomorphic to the circle. 

