$\operatorname{Aut}(S_4)$ is isomorphic to $S_4$

Solution 1:

Assuming that you know that $f$ fixes every $3$-cycle, it follows that $f$ also fixes every even permutation (because $A_4$ is generated by $3$-cycles).

Now $A_n$ is the commutator subgroup of $S_n$ when $n \geq 3$ (proof: write a $3$-cycle as the commutator of two transpositions). So the parity of an permutation is invariant under an automorphism of $S_n$.

Thus $f$ permutes the transpositions $(12), (13), (14), (23), (24), (34)$.

Now $(123) = f(123) = f((13)(12)) = f(13)f(12)$ so $f(12)$ is $(12)$ or $(13)$ or $(23)$.

Also $(12)(34) = f((12)(34)) = f(12)f(34)$ so $f(12)$ is $(12)$ or $(34)$.

Thus $f(12)$ is $(12)$ hence $f = 1$.