Showing that an automorphism of $S_4$ fixing each Sylow 3-subgroup must be the identity. I am working on a Hungerford exercise and trying to show that $S_4$ is isomorphic to its automorphism group. I know that there are four Sylow 3-groups in $S_4$ and the four Sylow groups exhaust the 3-cycles of $S_4$. Now, denote the 4 Sylow groups by $P_1$, $P_2$, $P_3$, $P_4$. If $f$ is an automorphism of $S_4$ that sends $P_i$ to $P_i$ for each $i$, then I want to show that $f$ must be the identity map of $S_4$. However it seems trickier than I thought and I cannot find a way to prove it. Could anyone please help me how to show that such $f$ must be the identity?
 A: As for proving $S_4\cong\operatorname{Aut}(S_4)$, recall that the automorphisms send conjugacy classes to conjugacy classes and preserve elements' order. In $S_4$, distinct conjugacy classes have elements of distinct orders except the conjugacy classes of the transpositions and the double transpositions (both whose elements have order $2$); but the former has size $6$, whilst the latter has size $3$, so no one automorphism can swap them. Therefore, every automorphism is cycle type-preserving, and then inner: $\operatorname{Aut}(S_4)=\operatorname{Inn}(S_4)$. But $\operatorname{Inn}(S_4)\cong S_4$, because $S_4$ is centerless. So, finally, $S_4\cong\operatorname{Aut}(S_4)$.
A: If $f\in Aut(S_4)$, first note that $f$ fixes (setwise) the set of all $3$-cycles as $f$ preserves elements of order $3$. Thus $f$ fixes $A_4$ as $A_4$ is generated by $3$-cycles. Therefore $f$ preserves the parity of the permutations.
If $\sigma$ is a transposition, then $f(\sigma)$ is an odd permutation of order $2$, hence a transposition as well. Let $P_1=\langle(123)\rangle$, $P_2=\langle(124)\rangle$, $P_3=\langle(134)\rangle$ and $P_4=\langle(234)\rangle$. Consider conjugations by transpositions of $P_i$'s given in the table:
$$\begin{array}{c|cccc}
&P_1&P_2&P_3&P_4\\\hline 
(12)&P_1&P_2&P_4&P_3\\
(13)&P_1&P_4&P_3&P_2\\
(14)&P_4&P_2&P_3&P_1\\
(23)&P_1&P_3&P_2&P_4\\
(24)&P_3&P_2&P_1&P_4\\
(34)&P_2&P_1&P_3&P_4
\end{array}$$
Since $(12)$ conjugates $P_3$ to $P_4$, $f((12))$ conjugates $f(P_3)=P_3$ to $f(P_4)=P_4$, so the only possibility for $f((12))$ (by looking in the table) is $(12)$. Similarly,  $f$ fixes every transposition. Since transpositions generate $S_4$, $f=id$.
