In a paper I read that $F_2\ltimes F_2^{2n-4}$ is a subgroup of $\mathrm{Aut}(F_n)$. The proof of this fact is as follows:
Choose $F_2\leq \mathrm{Aut}(F_2)$ and let it act diagonally on $F_2^{2n-4}$, i.e. the action is $\varphi:F_2\times F_2^{2n-4}\to F_2^{2n-4}$ given by $\varphi(u,(v_1,\dots,v_{2n-4}))=(uv_1,\dots,uv_{2n-4})$. This induces a semidirect product $F_2^{2n-4}\rtimes_{\varphi}\ F_2$. Now, to prove that this is a subgroup of $\mathrm{Aut}(F_n)$ we define the following map: \begin{align*} F_2^{2n-4}\rtimes_{\varphi}\ F_2&\rightarrow \mathrm{Aut}(F_n)\\ \mathbb{u}=((v_3,w_3,\dots,v_n,w_n),\alpha)&\mapsto \phi_\mathbb{u} \end{align*} where, if $F_n=\langle x_1,\dots,x_n\rangle$ we have $\phi_\mathbb{u}(x_i)=\alpha(x_i)$ for $i=1,2$ and $\phi_\mathbb{u}(x_i)=w_ix_iv_i^{-1}$ for $i\geq 3$, were we are assuming that $\alpha$ is an automorphism of $F_2$. We have to show that this is an injective map and then we will be done, but this is where I'm stuck.
Let $\mathbb{u}_1=((v_{1,3},w_{1,3},\dots,v_{1,n},w_{1,n}),\alpha_1)$ and $\mathbb{u}_2=((v_{2,3},w_{2,3},\dots,v_{2,n},w_{2,n}),\alpha_2)$ and suppose that $\phi_{\mathbb{u}_1}=\phi_{\mathbb{u}_2}$. Then $\alpha_1(x_i)=\alpha_2(x_i)$ for $i=1,2$ and so $\alpha_1=\alpha_2$. However, I don't know how to proceed with the second part, i.e. if $w_{1,i}x_iv_{1,i}^{-1}=w_{2,i}x_iv_{2,i}^{-1}$ for $i\geq 3$ I'm not sure how to prove that $w_{1,i}=w_{2,i}$ and $v_{1,i}=v_{2,i}$.
Thanks for your help.