Symmetric Group $S_n$ is complete I am just checking some properties of the symmetric group and found on Wikipedia the statement that "Conversely, for $n \neq 6$, $S_n$ has no outer automorphisms, and for $n \neq 2$ it has no center, so for $n \neq 2, 6$ it is a complete group, as discussed in automorphism group, below."
I could show that $Z(S_n)=1$ but do you know an easy proof for $\mathrm{Aut}(S_n)=\mathrm{Inn}(S_n)$ ?
 A: Lemma. Let $\mathrm{Aut}(G)$ be the automorphism group of a group $G$ and $\mathrm{cl}(G)$ the conjugacy classes of elements of $G$. Then $\mathrm{Aut}(G)$ acts on $\mathrm{cl}(G)$.
Explanation. Let $\alpha$ be an automorphism of $G$ and $\mathrm{cl}(a)=\{gag^{-1}:g\in G\}$ the conjugacy class of $a$. Then $\alpha(\mathrm{cl}(a))=\{\alpha(g)\alpha(a)\alpha(g)^{-1}:g\in G\}=\{t\alpha(a)t^{-1}:t\in G\}=\mathrm{cl}(\alpha(a))$ is also a class.
Lemma. If $\alpha\in\mathrm{Aut}(S_n)$ stabilizes the conjugacy class of transpositions then $\alpha\in\mathrm{Inn}(S_n)$ is inner.
Proof. We want to exhibit an element $\sigma\in S_n$ such that $\alpha(g)=\sigma g\sigma^{-1}$ for every $g$. It suffices to find a $\sigma$ for which $\alpha(\tau)=\sigma \tau\sigma^{-1}$ holds for every transposition $\tau$, since every permutation is a product of transpositions. It further suffices to find a $\sigma$ for which $\alpha(1k)=\sigma(1k)\sigma^{-1}$ for each $1< k\le n$, since these $n-1$ transpositions generate all the others.
Pick distinct $1<\ell_1,\ell_2\le n$. Then $\rho=(1\ell_2\ell_1)=(1\ell_1)(1\ell_2)$ has order $3$, in which case $\alpha(1\ell_1)\alpha(1\ell_2)$ has order three as well. Without loss of generality, $\alpha(1\ell_1)=(ab)$ and $\alpha(1\ell_2)=(ac)$ for some $a,b,c$; since this holds for each distinct pair $\ell_1,\ell_2$, we find that for every $k$, $\alpha(1k)=(af(k))$ for some $f(k)$.
Let $\sigma$ send $1$ to $a$ and $k$ to $f(k)$ for each $k>1$. This permutation satisfies our requirements.
Lemma. For $n\ne6$, the class of transpositions is the unique class of involutions with ${n\choose2}$ elements.
Proof. Consider the class of involutions with cycle type a product of $k>1$ disjoint cycles. Suppose
$$\frac{1}{k!}{n\choose 2}{n-2\choose2}\cdots{n-2(k-1)\choose2}={n\choose2}.$$
The LHS is the size of the new conjugacy class. Setting $m=n-2$ and $k=r+1$, cancelling ${n\choose2}$s,
$${m\choose2}\cdots{m-2(r-1)\choose2}=(r+1)!$$
The LHS is telescoping. Multiply by $2^r/(2r)!$ and get
$${m\choose2r}=\frac{(r+1)!2^r}{(2r)!}=\frac{r+1}{(2r-1)!!}$$
For $r>2$ the RHS is not even an integer (it is $<1$). The case $r=1$ gives $m(m-1)=4$ with no solution in $m$. Finally $r=2$ gives $m(m-1)(m-2)(m-3)=24$ which has positive solution $4$.
Theorem. For $n\ne6$, $\mathrm{Out}(S_n)=1$.
Proof. Combine the three lemmas. When $n\ne6$, automorphisms permute conjugacy classes and in particular stabilize the transposition class (since the size of this class is unique among the classes), hence every automorphism is inner.
Remark. This proof is adapted from this sketch from Wikipedia and these notes.
