Segal uses "similar" to mean conjugate. A "class of similar elements" means a conjugacy class.

The gist of Segal's argument is that an automorphism of a symmetric group must take a transposition to an element of order 2, but there is noa priorireason to suppose that it will take a transposition to a transposition (and in fact the outer automorphism of S_6 takes a transposition to a product of three transpositions).

So suppose that an automorphism of S_n takes a transposition to a product of k disjoint transpositions. The conjugacy class of the transposition contains elements. The conjugacy class of the product of k disjoint transpositions contains elements (i.e. the number of ways of choosing k pairs from n objects).

Segal skips the calculation which shows that these numbers can only be equal for k=1, unless n=6. Here's how I did it.

If then . Divide both sides by (n-2k)!(2k-2)! and you get

. Now look at the possible values of k. We're not interested in k=1, so start with k=2. The equation then says , which has no solution for n. If k=3 then we get , which has the solution n=6. If then , so the equation has no solutions.

The rest of Segal's proof is ingenious but not too hard to follow. What he does not do is to provide an example of an outer automorphism of S_6. I found a very nice geometrical construction of that here. (I also came across this account of a recent series of lectures that J-P Serre has been giving at Harvard, on the numbers 2 to 8 seen from a (very!) advanced viewpoint. The lecture on 6 includes some comments on .)