Hatcher's Algebraic Topology. Exercise 2.2.42 Update: My attempts are moved to my answer.
This is Exercise 2.2.42 in page 159 of Hatcher's Algebraic Topology.

Let $X$ be a finite connected graph having no vertex that is the endpoint of just one edge, and suppose that $H_1(X; \Bbb Z)$ is free abelian of rank $n>1$, so the group of automorphisms of $H_1(X; \Bbb Z)$ is $GL_n(\Bbb Z)$.
Show that if $G$ is a finite group of homeomorphisms of $X$, then the homomorphism $\phi:G \to GL_n(\Bbb Z)$ assigning to $g:X \to X$ the induced homomorphism $g_*:H_1(X;\Bbb Z) \to H_1(X;\Bbb Z)$ is injective.
Show the same result holds if the coefficient group $\Bbb Z$ is replaces by $\Bbb Z_m$ with $m>2$.
What goes wrong when $m=2$?

Are there any hints or suggestions? Thanks for your time and effort!
 A: Replacing $X$ by $\bigvee_{n} S^1$ does indeed lose generality. This question is about homeomorphisms of $X$, and their induced maps on homology. While is true that $X$ is homotopy equivalent to $\bigvee_{n} S^1$, the homeomorphism groups of $X$ and $\bigvee_{n} S^1$ need not be isomorphic, and so their images under the injective homomorphisms $G_1 = \text{Homeo}(X) \to G_n(\mathbb Z)$ and $G_2 = \text{Homeo}(\bigvee_{n} S^1) \to GL_n(\mathbb Z)$ will not be equal. For example, in rank 2, the group of homeomorphisms of the $\theta$ graph has order $12$, whereas the group of homeomorphisms of the $8$ graph (which is the wedge of two circles) has order $8$.
Still, though, if you can make a proof which works for the special case $X = \bigvee_{n} S^1$, that would be a good start. The general case is harder, but perhaps you could go on to generalize the special case proof.
Here's two hints for $\bigvee_{n} S^1$, the first of which is an important point that you do not seem to be considering: 


*

*Given an oriented edge of $\bigvee_{n} S^1$, what does it represent in homology? 

*Given a homeomorphism $g$ of $\bigvee_{n} S^1$ that does not take each oriented edge to itself preserving orientation, what would (1) imply about the matrix $\phi(g) \in GL_n(\mathbb Z)$?

A: (1) 
$H_{1}(X; \mathbb Z)$ is of rank $n > 1$, so $X \simeq \bigvee_{n} S^{1}$. 
Consider $X = \bigvee_{n} S^{1}$ first. 
To show $\phi: G \to GL_{n}(\mathbb Z)$ is injective, suppose $g: X \to X$ is homeomorphism s.t. $\phi(g) = \text{id}$, then $g$ maps each $S^{1}$ to itself and fixes the wedge point $x_{0}$.
Let $f = g|_{S^{1}}: S^{1} \to S^{1}$, then $f$ fixes $x_{0}$ and $f_{*} = \text{id}$, so $f$ preserves the orientation. 
$G$ is finite group, so $f$ is of finite order and there exists a smallest positive integer $k$ s.t. $f{^k}=\text{id}$. 
Let $y \in S^{1}$, $f(y) \neq y$, then points $y, f(y), f^{2}(y), \cdots, f^{k}(y) = y$ are permuted in $S^{1}$ clockwise or counterclockwise since $f$ preserves the orientation. 
Arc between $f^{i}(y)$ and $f^{i+1}(y)$ is mapped by $f$ to the next one, and such arcs cover $S^{1}$, so one of these arcs contains $x_{0}$, but $f$ fixes $x_{0}$. Contradiction. 
Thus $f = \text{id}: S^{1} \to S^{1}$ and $g = \text{id}: \bigvee_{n} S^{1} \to \bigvee_{n} S^{1}$. 
(2) 
For general finite connected graph $X \simeq \bigvee_{n} S^{1}$ $(n \ge 2)$, there exists a vertex $x_{0}$ of valence $\ge 3$. 
$x_{0}$ belongs to different loops based at $x_{0}$, and $g$ maps loops to themselves and preserves the orientation, so $g$ fixes $x_{0}$, and the followings are the same as the situation for $\bigvee_{n} S^{1}$. 
(3) 
For coefficient group to be $\mathbb Z_{m}$, $\phi: G \to GL_{n}(\mathbb Z_{m})$. Suppose $g: X \to X$ is homeomorphism s.t. $\phi(g) = \text{id}$. 
If $m > 2$, then $g$ preserves the orientation in each loop since $-\overline{1} = \overline{m-1} \neq \overline{1}$. 
If $m = 2$, $g$ may reverse the orientation in some loop  since $-\overline{1} = \overline{1}$. 
