I'm trying to proof that a map $ \varphi: Aut(F_n) \to GL_n(\mathbb{Z})$ is a homomorphism but i can't exactly define which is the function to show that.
The map $ \varphi $ is a map with for any $ \alpha \in Aut(F_n) $ the $(i,j)$-th entry of the matrix $ \varphi(a) $ is the sum of exponents of the letter $x_j$ in the $ a(x_i)$.
$F_n$ denotes the non-abelian free group generated by $n$ elements
On
In general, if $\kappa \colon G \to G/K$ is natural projection where $K$ is characteristic in $G$, then the map $\kappa$ induces a homomorphism $\tilde{\kappa} \colon \mathop{Aut}(G) \to \mathop{Aut}(G/K)$ given by $$\tilde{\kappa}(\phi)(g K) = \phi(g)K$$ for every $g\in G$ and $\phi \in \mathop{Aut}(G)$.
By setting $G = F_n$ and $K = [F_n,F_n]$ you can check that the map you get exactly the one have described.
The easiest way to see this is to note that this map factors through an abelianization of $F_n$. Namely, if we quotient by the commutator subgroup then an element $x_{i_1}x_{i_2}\cdots x_{i_n}$ reduces to $x_1^{m_1}x_2^{m_2}\cdots x_n^{m_n}$, where $m_i$ is the sum of the exponents of $x_i$ in the expression. For any automorphism $\alpha:F_n\to F_n$ there is a corresponding automorphism $\alpha':\mathbb{Z}^n\to\mathbb{Z}^n$ where $\alpha'(x_i)$ is the element obtained by combining the powers of each generator. But $\alpha'$ is exactly the element of $\mathrm{GL}_n(\mathbb{Z})$ that $\alpha$ is mapped to.