Group automorphism of $S_4$ I want to prove that $\operatorname{Aut}(S_4) \cong S_4$.
I saw $\operatorname{Aut}(S_4)$ is isomorphic to $S_4$ but i was troubled.
I know $Z(S_4)=1$, and so $S_4 \cong \operatorname{Inn}(S_4)$. Thus it is sufficient to show that every automorphism of $S_4$ is inner.
Now suppose that ${ \rm Syl}_3(S_4)= \lbrace P_1,P_2, P_3, P_4 \rbrace$. Thus $\operatorname{Aut}(S_4)$ acts on ${ \rm  Syl}_3(S_4)$.
$\textbf{Watson}$ say that this action is as follow:
$(P_i,f)=f(P_i)=P_{\sigma_f(i)}$. I still dont understand this action. Also why every automorphism of $S_4$ is inner?
 A: To move this old question out of the unanswered list, here's an argument which works up to $n=5$.
In general, given a group $G$, the group $\operatorname{Aut}(G)$ acts on the set of the conjugacy classes of $G$, $\mathscr{C}:=\{\operatorname{cl}(a), a\in G\}$, via $\varphi\cdot \operatorname{cl}(a):=\varphi(\operatorname{cl}(a))$. Essentially, the proof that this is indeed an action boils down to proving that $\varphi(\operatorname{cl}(a))=\operatorname{cl}(\varphi(a))$. Since inner automorphisms fix each conjugacy class, we have that $\operatorname{Inn}(G)\le\operatorname{ker}\phi$, where $\phi\colon\operatorname{Aut}(G)\to S_\mathscr{C}$ is the homomorphism equivalent to the said action.
Now, if $G=S_n$, then $\operatorname{Inn}(G)=\operatorname{ker}\phi$, because the stabilizer of the conjugacy class of the transpositions is precisely equal to $\operatorname{Inn}(S_n)$ (see e.g. here). Therefore (first homomorphism theorem), $\operatorname{Aut}(S_n)/\operatorname{Inn}(S_n)\cong\phi(\operatorname{Aut}(S_n))$. If distinct conjugacy classes are "tagged" by the distinct orders of their elements, then the action is trivial, because any automorphism (order-preserving) is then class-preserving, and finally $\operatorname{Aut}(S_n)=\operatorname{Inn}(S_n)$. This happens for $n=3$, being the three distinct conjugacy classes of $S_3$ made of elements of order $1$ (the identity), $2$ (the three $2$-cycles) and $3$ (the two $3$-cycles). Therefore $\operatorname{Aut}(S_3)=\operatorname{Inn}(S_3)\cong S_3$. By relaxing such a "strong" sufficient condition ("each conjugacy class, an order of its elements"), we can encompass $n=4$ and $n=5$ cases, too. In fact, if $S_n$ has two conjugacy classes of elements of the same order, but their sizes are different, than again all the automorphisms are class-preserving (being $|\varphi(\operatorname{cl}(\sigma))|=|\operatorname{cl}(\sigma)|$), and hence again the action is trivial (whence again $\operatorname{Aut}(S_n)=\operatorname{Inn}(S_n)$). For $n=4$, the conjugacy classes of the transpositions and of the double transpositions (both whose elements have order $2$) have sizes $6$ and $3$, respectively, and hence no one automorphism can "swap" each other. Likewise, for $n=5$, the conjugacy classes of the transpositions and of the double transpositions have sizes $10$ and $15$, respectively, and the same conclusion as for $n=4$ does hold.
This simple approach can't support any further in the $n=6$ case, where we have conjugacy classes of elements of the same order, which have also the same size.
