$\operatorname{Aut}(A_4)\simeq S_4$

Every element of $S_4$ gives you an automorphism of $A_4$ by conjugation, since $A_4$ is normal in $S_4$. The elements of $S_4$ which act trivially by conjugation on $A_4$ are precisely the elements of the centralizer $C_{S_4}(A_4)$, which is trivial, hence every element of $S_4$ gives you a distinct automorphism of $A_4$.

Let $G=A_4$,

By the fact you mentioned, $\operatorname{Aut}(G)\leq S_4 $ and since $\operatorname{Inn}(G)\cong G/Z(G)\cong G\leq\operatorname{Aut}(G)$; we have $$A_4\leq\operatorname{Aut}(G)\leq S_4$$.

Since $A_4$ is only subgroup of $S_4$ with order $12$ it is characteristic in $S_4$ which means there is a homomorphism from $ \operatorname{Aut}(S_4)\cong S_4$ to $ \operatorname{Aut}(G)$ and since $ \operatorname{Aut}(G)$ has at least $12$ elements and $S_4$ has no normal subgroup of order $2$, we have isomorphism from $S_4$ to $ \operatorname{Aut}(G)$.