Can we define a natural isomorphism $\phi: S_4 \to {\rm Aut}(A_4)$? I was recently viewing $\operatorname{Aut}(A_4)\simeq S_4$. Why is ${\rm Aut}(A_4)$ isomorphic to $S_4$ has been asked quite a bit on this site -- and I understand the logic in the linked answer. 
However, what I'm wondering is, can we actually write down a natural, explicit (injective and surjective) homomorphism between these two groups ? It's a bit hard for me to imagine this -- since the automorphism group is a group of maps, while $S_4$ consists of permutations. How can we send elements in the automorphism group to $S_4$ in a natural way ? 
Thanks for your help. (=
 A: Just given the bare fact that $\operatorname{Aut}(A_4)\cong S_4$, without considering whatever proof was given, here is a way to construct such an isomorphism:
First note that $\operatorname{Aut}(S_n)\cong S_n$ for almost all $n$, the only exceptions being $n=2$ and $n=6$. For each $n$ we have a natural map
$$S_n\ \longrightarrow\ \operatorname{Aut}(S_n):\ \sigma\ \longmapsto\ (\tau\ \mapsto\ \sigma\tau\sigma^{-1}),$$
which sends each $\sigma\in S_n$ to the automorphism of $S_n$ given by conjugation by $\sigma$. Note that this map in fact defines a group homomorphism $G\ \rightarrow\ \operatorname{Aut}(G)$ for any group $G$.
Because $A_4\subset S_4$ is the unique subgroup of $S_4$ of order $12$, it is mapped to itself by every automorphism of $S_4$. This means the map above induces a group homomorphism
$$S_4\ \longrightarrow\ \operatorname{Aut}(A_4):\ \sigma\ \longmapsto\ (\tau\ \mapsto\ \sigma\tau\sigma^{-1}).$$
Since you already know that $\operatorname{Aut}(A_4)\cong S_4$, to prove that it is an ismorphism, it suffices to prove it is injective, or equivalently, that only the identity element $e\in S_4$ acts trivially on $A_4$ by conjugation. Put more simply; that the only element $\sigma\in S_4$ for which $\sigma\tau\sigma^{-1}=\tau$ holds for all $\tau\in A_4$, is the identity $e\in S_4$.
