Assume that there is a transitive subgroup $H$ of $S_n$ w.r.t. the standard action $S_n\curvearrowright \{1,2,\cdots, n\}:=X$ such that $H\cong S_k$. Is there any sharp estimate on the upper bound of $n$ in terms of $k$ for large $n$, say for all $n\geq 5$?
For example, one can show that $S_n=HK$, where $K$ is the stabilizer subgroup of $S_n$ for any point in $X$ and hence $K\cong S_{n-1}$. Hence we know $n!\leq k!\cdot (n-1)!$, thus, $n\leq k!$. In particular, is this estimate sharp for $n\geq 5$?
Thanks!
Here is a summary of the answers in the comments.
As runway44 says, the answer to the question asked is yes, $k!$ is a sharp upper bound, because the regular representation of $S_k$ is transitive on $n=k!$ points.
A possibly more interesting question would be to as for a lower bound for $n$ in terms of $k$ with $n>k$. This is $k(k-1)/2$ for $k>2$ and $k \ne 5$. For $k=5$ we can have $n=6$.