Let $G$ be a finite group and $p$ a prime divisor for the order of $G$. Show that if $p^{2}$ divides the order of G, then $p$ divides $|Aut(G)|$ .
I know that $G$ has a subgroup of order $p$ and I think that it would be a good idea to consider the following homomorphism to prove that $Aut(G)$ has a subgroup of order $p$ : $$ \phi :G \to Aut(G)$$ $$\phi(g) = C_{g}$$ where $C_{g}$ is the conjugation given by $g$. Also I know that $Ker(\phi)=Z(G)$ then $G/Z(G) \cong Im(\phi)=Inn(G)$, but I can't figure out how to use the fact that $p^{2}$ divides the order of $G$.