Let $\Gamma$ be a finite graph, then $H^1(\Gamma,\mathbb{Z})\cong \mathbb{Z}^{g(\Gamma)}$ can be viewed as a $\mathrm{Aut}(\Gamma)$ module. 


Conversely, given a finite group $G$, and a $G$-module $\mathbb{Z}^n$, does there always exist a finite graph $\Gamma$ such that the $G$ module $\mathbb{Z}^n$ arises as $G\overset{f}{\to}\mathrm{Aut}(\Gamma)\curvearrowright H^1(\Gamma,\mathbb{Z})$ for some $f\colon G\to \mathrm{Aut}(\Gamma)$?



Finite group representation as $\mathrm{Aut}(\Gamma)$ action $H^1(\Gamma,\mathbb{Z})$ of graph?
No. An action of a group $G$ on a graph $\Gamma$ induces a homomorphism $G\to \mathrm{Out}(\pi\_1(\Gamma))=\mathrm{Out}(F\_n)$. So a representation $G\to \mathrm{GL}\_n({\mathbb Z})$ can come from an action on a graph only if it lifts to a homomorphism $G\to \mathrm{Out}(F\_n)$ over the quotient homomorphism $\mathrm{Out}(F\_n)\to \mathrm{GL}\_n({\mathbb Z})$. [This paper of Zimmermann](https://arxiv.org/pdf/math/0604464.pdf) gives an example of a finite cyclic subgroup of $\mathrm{GL}\_n({\mathbb Z})$ that does not lift.


