Show that for \(n\ne6\), \[ \operatorname{Aut}(S_n)=\operatorname{Inn}(S_n). \]
Solution
For \(n=1,2\) the statement is immediate, so assume \(n\ge3\).
1. Every automorphism sends transpositions to transpositions when \(n\ne6\).
An involution in \(S_n\) has cycle type \(2^k1^{n-2k}\). Its conjugacy class has size \[ \frac{n!}{2^k k!(n-2k)!}. \] A transposition has class size \(\binom n2\). Automorphisms preserve element order and conjugacy-class size, so the image of a transposition must be an involution satisfying \[ \frac{n!}{2^k k!(n-2k)!}=\binom n2. \] For \(k\ge2\) this is equivalent to \[ \frac{(n-2)!}{2^{k-1}k!(n-2k)!}=1. \] For \(k=2\) there is no integer solution. For \(k=3\) the unique solution is \(n=6\). For \(k\ge4\), already at the smallest possible \(n=2k\) the left side is \[ \frac{(2k-2)!}{2^{k-1}k!}>1, \] and it increases with \(n\). Thus, when \(n\ne6\), necessarily \(k=1\).
2. An automorphism preserving transpositions is inner.
Let \[ s_i=(i\ i+1),\qquad 1\le i<n. \] These generate \(S_n\). Put \(t_i=\phi(s_i)\). Each \(t_i\) is a transposition. Moreover, \[ \operatorname{ord}(t_it_j)= \operatorname{ord}(s_is_j) = \begin{cases} 3,&|i-j|=1,\\ 2,&|i-j|>1. \end{cases} \] Two distinct transpositions have product of order \(3\) exactly when their supports meet in one point, and they commute exactly when their supports are disjoint.
Write \(t_1=(a_1a_2)\). Since \(t_2\) meets \(t_1\) in exactly one point, after interchanging \(a_1,a_2\) if necessary write \(t_2=(a_2a_3)\). Inductively, \(t_i\) must meet \(t_{i-1}\) but be disjoint from \(t_1,\ldots,t_{i-2}\), so \[ t_i=(a_i a_{i+1}) \] for distinct \(a_1,\ldots,a_n\).
Let \(\pi\in S_n\) be defined by \(\pi(i)=a_i\). Then \[ \pi s_i\pi^{-1}=t_i=\phi(s_i) \] for every \(i\). Since the \(s_i\) generate \(S_n\), \[ \phi(g)=\pi g\pi^{-1} \qquad(g\in S_n). \] Hence \(\phi\) is inner.
Therefore \[ \operatorname{Aut}(S_n)=\operatorname{Inn}(S_n) \] for all \(n\ne6\).