On what sets can $\mathfrak{S}_n$ act transitively? I would like to know $\mathfrak{S}_n$ could act faithfully transitively on sets with $m$ elements, with $m > n$.
I know that it is not possible if $m = n+1$ except for $n = 5$.
Any ideas ?
 A: There are many sets of more than $n$ elements on which $\mathfrak S_n$ acts transitively. The largest possible example is that of the $n!$ total orderings of the set of $n$ elements (the one used to define $\mathfrak S_n$). One can deduce numerous smaller examples from this one.
A: There is a complete classification of sets equipped with a transitive $G$-action. Namely:

If $G$ acts on a set $X$ (on the left) transitively, then there is a subgroup $H \le G$ and a $G$-equivariant bijection between the set $[G : H]$ of (left) cosets of $H$ (with the evident (left) $G$-action) and $X$.

In fact, this is just a different way of stating the orbit–stabiliser theorem. Thus, $G$ can act transitively on a set of $n$ elements if and only if $G$ has a subgroup of index $n$.
A: This question is equivalent to asking what the conjugacy classes of subgroups of $S_{n}$ are, since each transitive permutation representation of any finite group is permutation equivalent to the action on the cosets of one of its subgroups, and conjugate subgroups lead to equivalent permutation representations. Since every finite group embeds in some symmetric group, this would soon become an enormous task without further information.
For larger $n,$ I think that the next smallest degree of a faithful permutation representation of $S_{n}$ ( after $n$) is $\frac{n(n-1)}{2},$  which comes from  the action on unordered pairs from $\{1,2, \ldots n \}.$ There are small exceptions to this though.
