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I was wondering if there is an easy counter example to what follows:

Suppose that $E$ is contractible CW-complex and $G_{1}, G_{2}$ are two isomorphic groups acting freely and continuously on $E$.

Is it true that the two actions are conjugated ?

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I may be misunderstanding the question (namely, the use of 'conjugated'), but whay about $\mathbb Z$ acting on $\mathbb R$ by $x\circ n=x+n$ and $x\bullet n=x-n$? – Fernando Muro Apr 29 '14 at 16:25
@FernandoMuro: My understanding of the question may be different to yours, but aren't they conjugate by $x\mapsto -x$? But what about $\mathbb{Z}$ acting on a copy of $\mathbb{R}$ with little blobs around each integer, by $x\mapsto x+n$ and $x\mapsto x+2n$? – Jeremy Rickard Apr 29 '14 at 16:36
@JeremyRickard Thanks, I thought that 'conjugated' meant twisted by inner automorphisms. – Fernando Muro Apr 29 '14 at 16:40
@FernandoMuro: Ah, right. So "conjugate by an element of $G$" versus "conjugate by an element of $\operatorname{Aut}(E)$". Yes, I guess both are reasonable interpretations of the question. – Jeremy Rickard Apr 29 '14 at 16:46

You can make a free group of rank two act on the plane (or the hyperbolic plane if you prefer) in two ways such that the orbit spaces are not homeomorphic: one is a punctured torus and the other is a three times punctured sphere.

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Thank you Tom! I was wondering if we add the assumption that $E/G_{1}$ and $E/G_{2}$ are compact do you still have an other counter example?! – Ilias Amrani Apr 29 '14 at 17:27
@Fedotov, if you assume that E is a contractible manifold, that is very nearly the (open) Borel conjecture: if two aspherical compact manifolds of the same dimension are homotopy equivalent, then they are homeomorphic. Here, asphericity implies that homotopy equivalence is the same as having the same fundamental group, which is true of your setup. – Craig Westerland Apr 29 '14 at 17:54
Dear Craig, I wanted to add this ad the begining of the question. But I think that the question is little bit different since I assume in the hypothesis that the universal cover is the same (or homemorphic) and I dorp the hypothesis to be a manifold. I just asked Tom what happens if $E/G_{1}$ and $E/G_{2}$ are assumed to be compact. – Ilias Amrani Apr 29 '14 at 18:00
I agree that it is a different question. I'm just pointing out that if Tom has a counterexample when the $E/G_i$ are compact manifolds, then he also has a counterexample to the Borel conjecture. I certainly would be excited to hear it. Tom? – Craig Westerland Apr 29 '14 at 18:12
I don't think anybody is going to disprove any big conjectures on this thread! Here's a compact 1-dimensional non-manifold example. The graph that looks like the letter $\theta$ and the graph you get by connecting two circles with an arc have isomorphic fundamental groups and homeomorphic universal covers. – Tom Goodwillie Apr 29 '14 at 20:12

The example I gave in comments has $E/G_1$ and $E/G_2$ compact (but not manifolds), at least if you use compact "blobs".

More explicitly, let $E$ be the subset of $\mathbb{R}^2$ given by the union of the $x$-axis and closed discs of equal radius around $(n,0)$ for every integer $n$. Let $G_1=\mathbb{Z}$ acting by integer translations, and $G_2=\mathbb{Z}$ acting by even integer translations.

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Now, I see what you meant! Very nice example! – Ilias Amrani Apr 29 '14 at 18:42

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