TITLE: How many automorphisms of $S_n$ take transpositions into transpositions?
QUESTION [13 upvotes]: I need to show that an automorphism of $S_n$ which takes transpositions to transpositions is an inner automorphism.
I thought it could be done by showing that such automorphisms form a subgroups $H\le Aut(S_n)$, that $Inn(S_n)\subset H$ and that they have the same number of elements. The number of inner automorphisms is $n!$ because $S_n$ has a trivial center (at least when it is not abelian) and therefore is isomorphic to $Inn(S_n)$. However I have no idea how I could count the number of elements in $H$.
Is there a way to do it, or should I change the approach altogether?
Thank you.

REPLY [3 votes]: I think there is an easier way of looking at this (now edited to get the argument correct - and also simpler, thanks to Jyrki's comment). An inner automorphism of $S_n$ is equivalent to a permutation of the underlying set on which $S_n$ acts. Let's choose our generators to be a permutation $a=(1 2)$ and the n-cycle $b=(1 2 3 ... n)$.
Then we know that the automorphism sends $a$ to another transposition $a'$. 
Consider the cycle decomposition of $b'$, the image of $b$. The two objects permuted by $a'$ must be adjacent within the same cycle in $b'$ (else $b'^{-1}a'b'$ will be found to be a transposition moving different objects from $a'$ and will commute with it, and this is not true for $a, b$). 
Suppose the objects moved by $a'$ are adjacent in a cycle of length $m&ltn$ in $b'$. Then $b'^m$ will commute with $a'$ (it will not move the objects moved by $a'$). This is not true for $a, b$, so we must have $m=n$ i.e. $b'$ is an $n$-cycle with the elements moved by $a'$ adjacent. Since this is the same cycle pattern as for $a, b$ there will be a permutation of the underlying objects which does the job, and this is the required inner automorphism.