How much does $\operatorname{Aut}(H_1(S))$ determine a homeomorphism $S \to S$? Let $S$ be an orientable compact surface. A homeomorphism $f: S \to S$ induces an isomorphism $f_{*}: H_1(S) \to H_1(S)$.
How much can we say the converse? Namely, if we are given an element of $\alpha \in$ $\operatorname{Aut}(H_1(S))$, is there a self-homeomorphism $f$ of $S$ (unique up to isotopy or something) such that $f_*=\alpha$?
 A: The set of self-homeomorphisms of a surface up to isotopy is called the mapping class group of the surface. For genus one, this is determined by the action on homology, but for higher genus there is a very large and interesting subgroup of the mapping class group, called the Torelli group, which consists of self-homeomorphisms inducing the trivial map on homology. 
A: The answer depends on what you mean by $\mathrm{Aut}(H^1(S))$. If you mean general linear group, then the answer is no, but if you mean the symplectic group, then the answer is yes.
There is a cup product pairing $H^1(S) \times H^1(S) \to H^2(S) \cong \mathbf Z$ which is symplectic, and any automorphism of the surface preserves this pairing (up to a sign, if you don't require it to preserve the orientation of $S$). This is the only condition: any symplectic automorphism can be realized by a homeomorphism of the surface.
See also http://en.wikipedia.org/wiki/Mapping_class_group#Torelli_group
A: Here's an example without the assumption that $S$ is a surface:
Let $S = S^1 \vee A$ where $A$ is a closed annulus. Then $H_1(S) = \mathbb Z \oplus \mathbb Z$. Take the isomorphism $\varphi: (a,b) \mapsto (b,a)$.
A homeomorphism $f$ inducing $\varphi$ would have to map $x$ in $S^1$ to $f(x) \in A$. But $S^1$ and $A$ are not homeomorphic.
