Suppose that we have two closed nmanifold $M$ and $N$ such that the topological group of homeomorphisms $Homeo(M)$ is homotopy equivalent to $Homeo(N)$ (maybe as topological groups if needed), can we deduce that $M$ is homotopy equivalent (or even homeomorphic ) to $N$ ? Is there an easy counterexample ?
It is a result of Whittaker (1963), and Rubin (1990, in greater generality with a better proof) that $Homeo(M)$ (viewed as a group) determines $M$ up to homeomorphism (see this question). I doubt that the homotopy type is enough: Gabai had shown that for closed hyperbolic 3manifolds, the inclusion of $Isom(M)$ into $Diff(M)$ is a homotopy equivalence. $Isom(M)$ is a finite group, so its homotopy type is given by its order. I don't think it is hard to find two hyperbolic manifolds with the same cardinality of isometry group (in fact, I assume that order is usually equal to $1.$)

$\begingroup$ Thanks for your answer and reference! I just edited my question, missed one word "groups" which is crucial. In this case I will be happy to see your counterexample for closed hyperbolic 3maniflods. $\endgroup$ – google Jul 20 '15 at 1:25

5$\begingroup$ There are lots of distinct hyperbolic 3manifolds with trivial symmetry group. You can find these using SnapPy and even verify the numerical computations rigorously. For a concrete example, take surgeries on the KinoshitaTerasaka knot 11n42 with very high coefficient. The knot complement itself has trivial symmetry group, and this property is preserved by very high surgeries. $\endgroup$ – Danny Ruberman Jul 20 '15 at 2:02

$\begingroup$ @DannyRuberman I'm missing something, the complement is hyperbolic 3manifold, is it closed ? $\endgroup$ – google Jul 20 '15 at 2:09

1$\begingroup$ The Dehn surgered manifolds are closed manifolds. $\endgroup$ – ThiKu Jul 20 '15 at 4:13

1$\begingroup$ @google To elaborate on ThiKu's comment: Dehn surgery means adding a solid torus via some identification of the boundary, encoded in a pair of integers. Thurston's Dehn Surgery theorem says that for all but finitely pairs, you get a closed hyperbolic manifold. Moreover, the core of the solid torus, for large integer pairs, is the shortest closed geodesic, and so is preserved by any isometry. Hence it gives an isometry of the knot complement, so if that has trivial symmetry group, so does the surgered manifold. $\endgroup$ – Danny Ruberman Jul 20 '15 at 12:58