This question is related to, but much more specific than, this one.

For $k \leq n$, let $a(k,n)$ denote the number of conjugacy classes of subgroups of the symmetric group $S_n$ which are isomorphic to $S_{n-k}$. It is straightforward (see this argument by Christopher Ryba), but not trivial, to show that for $n \neq 6$ we have $a(1,n)=1$.

It is easy to see that $a(2,n) \geq 2$, since we have the standard embedding $S_{n-2} \subset S_n$ as well as the one which sends odd permutations $\sigma$ to $\sigma (n-1 \: n)$. More generally, by adding some transpositions to the odd permutations, we get that $a(k,n) \geq \lfloor k/2 \rfloor +1$.

Weird things happen to the sequences $a(k,n)_{n=1,2,3,...}$ when $n-k \leq 5 < n$ because of the exotic copy of $S_5$ in $S_6$, but it seems that they may stabilize after this.

**Question:** Does $A(k):=\lim_{n \to \infty} a(k,n)$ exist for all $k$? Is there a reasonable formula (maybe just $\lfloor k/2 \rfloor +1$)?