Can I get ${\rm{Inn}}(S_3)={\rm{Aut}}(S_3)$ from $\varphi\in{\rm{Aut}}(S_3)$ being equivalent to a faithful transitive action of $S_3$ on $\{1,2,3\}$?

Question

An automorphism of $S_3$, say $\varphi$, is equivalent to a faithful action of $S_3$ on the set $X:=\{1,2,3\}$. The orbit equations $3=1+1+1$ and $3=1+2$ are ruled out, because they would both imply $\operatorname{ker}\varphi=\bigcap_{i\in X}\operatorname{Stab}(i)\ne\{()\}$. Therefore, the only option left is this action to be also transitive. To sum up, this action has stabilizers which:

1. trivially intersect;
2. are conjugate in $S_3$.

Can I use 1 and 2 to prove that $\varphi$ is inner?

Answer 1

Apparently, an answer is desired. The answer is certainly 'no', because everything written in the question is also true for $S_6$, where not every automorphism is inner (and other permutation groups as well). Thus the two properties cannot be used to prove that every automorphism is inner.

This was why all of the comments are talking about $S_6$ (or $M_{12}$, which also has an outer automorphism).

More prosaic than the outer automorphism of $S_6$, the group $A_3$ also satisfies the conditions, and obviously has an outer automorphism.

Answer 2

It seems to me that the answer is "yes, one can, although you haven't finished doing so." It is indeed true that an automorphism of $S_n$ is a map $\phi : S_n \to S_n$ and thus can be thought of as an action of $S_n$ on $\{ 1,2, \ldots, n \}$.

It is also true that this action will necessarily be faithful and transitive: If it were not faithful, then $\phi$ would have a kernel, contrary to the claim that $\phi$ is an automorphism and, if it were not transitive, then the image of $\phi$ would land in $S_{a_1} \times S_{a_2} \times \cdots \times S_{a_k}$, where $a_j$ are the orbit sizes, contradicting that $\phi$ is an automorphism.

And, as you say, this implies that there is a subgroup $H$ of $S_n$ with index $n$ and $\bigcap_{g \in S_n} gHg^{-1} = \{ e \}$.

On the other hand, all of this is correct for arbitrary $n$ and, when $n=6$, then $PGL_2(\mathbb{F}_5)$, acting on the projective line $\mathbb{P}^1(\mathbb{F}_5)$, is a subgroup $H$ with the required properties; this gives rise to the outer automorphism of $S_6$. So you are not done, you actually need to rule out such an $H$.

Of course, the group $S_3$ is small enough that you can simply list all index $3$ subgroups of $S_3$ by hand, and perhaps that is what you intended to do. But something like that is what you need to do to finish the proof.