There are a lots of questions (and several good explanations) on understanding the non-trivial outer automorphism of $S\_6$. What I haven't seen, though, is a good explanation of why there are no such outer automorphisms for $n\neq 6$. Wikipedia makes a passing mention of this fact [here](http://en.wikipedia.org/wiki/Automorphisms_of_the_symmetric_and_alternating_groups#No_other_outer_automorphisms).

In particular, they have this cryptic claim:

> 
> For every symmetric group other than $S\_6$, there is no other conjugacy class of elements of order 2 with the same number of elements as the class of transpositions.
> 
> 
> 

This seems like the natural thing to prove, given the knowledge that you can construct an outer automorphism of $S\_6$ which takes transpositions to products of three transpositions (i.e. things like $(12)\mapsto (12)(34)(56)$.) Also, the second reason Wikipedia gives doesn't feel very intuitive to me.

**My question, then, is how does one justify the original (quoted) statement above?**

Here is the naive computation of sizes of conjugacy classes: We need to show that when $n, k\neq 6,3$
$$\frac{n!}{2^k k! (n-2k)!}\neq\frac{n!}{2(n-2)!},$$which is equivalent to $$2^k k!(n-2k)!\neq 2(n-2)!$$

What I want to say is something like, if $n{-}2$ is bigger than $k$, then there is always some factor on the RHS that doesn't appear in the LHS (except when it is a power of two). Will this type of analysis get me anywhere, or is there a better way to think about this?