$S_n$ acts faithfully on a finite set $X$ implies an orbit of size $n$

Question

If $S_n$ acts faithfully on a finite set $X$, must there be an orbit of size $n$?

The context of the question is the following: suppose a Galois extension $E/F$ has Galois group $G$ isomorphic to the symmetric group $S_n$. Let $f \in F[x]$ be a separable polynomial of degree $m$ with splitting field $E$, and consider the action of $G$ on the roots of $f$. Does there exist a Galois orbit of $n$ roots of $f$ and thus an irreducible factor of $f$ of degree $n$?

We can rephrase it as the following: if $m \geq n$, then $S_m$ has subgroups isomorphic to $S_n$. Is the behavior of this subgroup necessarily just $S_n$ restricted to various $n$ element subsets, fixing the rest of the points?

Answer 1

No: consider the left Cayley action of $S_n$ on itself, i.e. $g\circ h = gh$. This action is clearly faithful (and in fact redundantly so) and it's also transitive, since for any $h$ we can act by $gh^{-1}$ to take it to any $g$ we like. Thus there is a single orbit of size $n!$.