I am to find the number of isomorphisms between $S_n \to S_n$. How do I proceed? One is the map which maps all the elements to itself.
So if $\varphi(ab)\to (cd)$ will be one case. Consider the case whe $n=20$. Now $\varphi(ab) \to (cd)(ef)$ because the number of order 2 cycles is $20_{C_{2}}$ and the number of 2 2-cycles is $(20_{C_{2}}×18_{C_{2}})/2$. Now if $\varphi$ is a isomorphism then $\varphi^{-1}$ is also an isomorphism so if $190$ elements are mapped to $270$ then $270$ should also be mapped to some $190$ elements.
Well I am really not able to conclude why this will not happen.
Allow me to spill the beans.
An isomorphism of a group with itself is called an automorphism. The set of all such forms a group, denoted $\rm{Aut}(G)$, with function composition as the operation.
Interestingly enough, except for $S_2$ and $S_6$, we have $\rm{Aut}(S_n)\cong S_n$.
We know, for $n\gt2$, that $S_n$ is center less. Thus it embeds as a subgroup of its automorphism group.
With a little work, one can show that all the automorphisms are inner ($n\ne6$). This means $S_n$ is (in fact equal to) the automorphism group.
Inner automorphisms are the ones of the form $i_g(x)=gxg^{-1}\,,\forall x$.
That's all for now.