I have to prove that $\operatorname{Aut}(A\_4)\simeq S\_4$.

If $G$ is a finite group, $p$ divides $|G|$, $n=n\_p(G):=|\operatorname{Syl}\_p(G)|$ then we know that $\operatorname{Aut} G\le S\_n$. I used this to prove that $\operatorname{Aut}(A\_4)\le S\_4$, since $n\_3(A\_4)=4$.

But now, how can I show the inverse inclusion, i.e. that $S\_4\le \operatorname{Aut}(A\_4)$?

I know that $S\_4\simeq \operatorname{Aut}(S\_4)$ but I don't know if it could help.

Any hint would be appreciated! Thanks!