$\textbf{Exercise.}$ If $G=\langle x,y ~|~ x^2=1=y^{2^n},y^x=y^{1+2^{n-1}} \rangle$, prove that $\mathrm{Aut}(G)$ is a $2$-group.
I tried using this result, calculated the Frattini subgroup of $G$ and found this to be $\langle y^2 \rangle$. Then $|\mathrm{Aut}(G)|$ divides $3.2^{2n-1}$, but I can't show why $3$ doesn't divide the order of $\mathrm{Aut}(G)$.
An automorphism of order $3$ would have to cycle the three non-trivial cosets of the Frattini subgroup $\Phi(G)$. So, without loss, suppose it carries $x$ into the coset $y\Phi(G)$. As $x$ has order $2$ we would have that its image has order $2$, that is $x=y^{2^{n-1}}\in\Phi(G)$. But clearly the automorphism (and its inverse) must centralize the cyclic group $\Phi(G)$. This is a contradiction.