How to show that an automorphism of $S_n$ is inner? Given an automorphism $\phi:S_n\rightarrow S_n$ such that it maps all the transpositions on the transpositions, how do I show that this map is given by a conjugation with an element $s\in S_n$?
Thanks in advance!
 A: Let's suppose that we are looking at $\phi:\ S_N \to S_N$ with $N \geq 3$; for small $N=1,2$ this is trivial.
Let $\phi( (n,m) ) =: \sigma_{n,m}$. We can notice that:
$$\sigma_{n,m} := \phi((n,m))  = \phi((1,n)(1,m)(1,n)) = \sigma_{1,n} \sigma_{1,m}\sigma_{1,n}.$$
Now, because $\phi$ was supposed to be bijective, all the $\sigma_{n,m}$ are different for different pairs of $n,m$. In particular, if $\sigma_{1,n} = (a_n,b_n)$, then $\{a_n,b_n\} \cap \{a_m,b_m\} \neq \emptyset$ for any $n,m \neq 1$, $n \neq m$. A simple combinatorial argument shows the all the pairs $\{a_n,b_n\}$ have an element in common. Indeed, we may assume without loss of generality that $a_2 = a_3 =:a$ and $b_2 \neq b_3$. If $N =3$ we are done. Else for all $m \geq 3$ we have two options: either $a \in \{a_m,b_m\}$ (and we may wlog say that $a_m = a$) or $\{a_m,b_m\} = \{b_2,b_3\}$. If $N \geq 5$, the second case can never happen. If $N = 4$, the second case might a priori happen. Allowing $a_2 = a_3 = 2,\ b_3 = b_4 = 3,\ b_2 = a_4 = 4$, we find that $\phi$ would have to act as follows: $(12) \mapsto (24),\  (13) \mapsto (23),\ (14) \mapsto (34)$. But this quickly leads to a contradiction, because for instance $(12)(13)(14)$ is a $4$-cycle, while $(24)(23)(34)$ is a $3$-cycle.
Thus, wlog $a_n = a$ for all $n$ for some $a \in [N]$. 
Consider the permutation $\pi$ which maps $1$ to $a$, and $n$ to $b_n$. Clearly, conjugation by $\pi$, say $\gamma_\pi$, sends $(1,n)$ to $$\gamma_\pi((1,n))= \pi(a_n,b_n) = (a,b_n) = \phi((1,n)).$$ Thus, $\phi$ and $\gamma_\pi$ agree on all transpositions of the form $(1,n)$. But they are both automorphisms, and these transpositions generate $S_N$, so they are equal everywhere, and we are done.
