Hatcher's Algebraic Topology. Exercise 2.2.42
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 $\phi(g) \in GL_n(\mathbb Z)$?