I have a curiosity about homology spheres: I was wondering if they were uniquely characterized by their fundamental group. I.e. given two $n$dimensional (integral) homology spheres with isomorphic fundamental groups, are they homeomorphic? If not, how many homeomorphism classes corresponds to a given fundamental group?

8$\begingroup$ Pick an irreducible representative of an element $[M]$ of the homology cobordism group $\Theta_{\Bbb Z}^3$ that is not 2torsion. Then it does not support an orientationreversing homeomorphism. Examples are the Brieskorn spheres $\Sigma(p,q,pqk1)$ for $k \geq 1$ and $p,q$ relatively prime. Then $M \# M$ and $M \# \overline{M}$ provide counterexamples by prime decomposition. (If you demanded irreducible then your answer is of course 'yes' for 3manifolds because other than $\Sigma(2,3,5)$ and $S^3$, these are aspherical and the Borel conjecture is known in 3 dimensions). $\endgroup$ – Mike Miller Jul 9 '15 at 16:20
See edit at bottom for further information answering the question in all dimensions.
In all odd dimensions $2k 1 > 3$, there are nonhomeomorphic homology spheres with fundamental group G = the binary icosahedral group (fundamental group of the Poincaré homology sphere). This follows from basic surgery theory; the Wall group $L_{2k}(G)$ contains a subgroup of the form $\mathbb{Z}^n$ for some $n$ related to the number of irreducible complex representations of $G$. This subgroup is detected by the socalled multisignature, as described in Wall's book on surgery theory.
Choose one homology sphere $M^{2k1}$ with $\pi_1(M) = G$. For instance, you can repeatedly spin the Poincaré sphere, where spinning $P$ means doing surgery on the obvious $S^1$ in $S^1\times P$. For any $M$ with $\pi_1(M) = G$, there's an invariant originally described by Atiyah and Singer. A priori, it's a smooth invariant, but is known to be a homeomorphism invariant. Roughly speaking, one knows that for some $d \in \mathbb{N}$, the manifold $d\cdot M$ is the boundary of a $2k$manifold $X$ with $\pi_1(X) = G$. Then $X$ has a collection of equivariant signatures (associated to the action of $G$ on the universal cover of $X$), known collectively as the multisignature. The multisignature (divided by $d$) is a topological invariant. (Technically, this is only welldefined up to a choice of isomorphism $\pi_1(X) \to G$ but this is readily dealt with.)
Now, the group $L_{2k}(G)$ acts on the structure set of $M$, as described in Wall's book. The effect of the action is to change the multisignature, and hence it changes the homeomorphism type of $M$. In this construction, you not only preserve the fundamental group, but also the (simple) homotopy type of $M$.
It's possible that acting on an evendimensional homology sphere $M^{2k}$ with fundamental group $G$ by elements of $L_{2k+1}(G)$ could change the homeomorphism type. But odd dimensional $L$groups are much harder, and you'd need some serious expertise to see what the effect should be. I rather suspect that you could do something simpler, and change the homology with local coefficients or something like that.
Addendum: There are two papers of Alex Suciu that answer this question in dimensions at least 4. In "Homology 4spheres with distinct kinvariants," Topology and its Applications 25 (1987) 103110, he gives examples of homology 4spheres with the same $\pi_1$ and $\pi_2$ that are not homotopy equivalent. In "Iterated spinning and homology spheres", Trans AMS 321 (1990) he constructs homology nspheres in dimensions $n \geq 5$ with the same property. Combined with the remarks above about dimension 3, this answers your question.

$\begingroup$ Could, you, please have a look at my answer to see if it is correct? Thank you. mathoverflow.net/a/316175/121665 $\endgroup$ – Piotr Hajlasz Nov 25 '18 at 19:55
