I'd like to demonstrate that every automorphism of $S_3$ is a conjugation but without looking at the order of the elements. My attempt is the following :
I've already proved that there is an injective homomorphism $f : Aut(S_3) \hookrightarrow S_3$ and also that the conjugation map $Ad$ is an automorphism.
Now let's consider $g : S_3 \longrightarrow Aut(S_3)$ such as $g(\sigma) = Ad(\sigma)$. It is easy to prove that $g$ is an injective homomorphism.
Let $\psi$ be in $Aut(S_3)$ and $\sigma$ be in $S_3$. Thus we have :
$$g \circ f : Aut(S3) \hookrightarrow S_3 \hookrightarrow Aut(S_3)$$ $$ (g \circ f)(\psi) = g(\sigma) = Ad(\sigma)$$
Because of the two injections, we can deduce that $S_3 \cong Aut(S_3)$ and thus $g \circ f$ is an automorphism. I think it also proves that there is as much elements in $Aut(S_3)$ than conjugations by the elements of $S_3$. Therefore, as the set of conjugations by the elements of $S_3$ is a subset of $Aut(S_3)$ we can deduce that every automorphism of $S_3$ is a conjugation.
I'm not quite sure about what I said. Could you please help me ?
Thanks a lot.