Subgroups of $S_n$ of index $n$ are isomorphic to $ S_{n-1}$

I am trying to show that the subgroups of $S_n$ of index $n$ are isomorphic to $ S_{n-1}$, if $n\ge 2$.

I tried to do this as follows: Let $G < S_n$ be a subgroup of index $n$, and let it act on a set of cosets $S_n/G \setminus 1\cdot G$ by multiplication from left. (It is easy to check this is indeed an action.) Then this action defines a homomorphism $\phi:G\rightarrow\mathrm{Aut}_{\mathrm{Set}}(S_n/G \setminus 1\cdot G) \cong S_{n-1}$. One needs to show $\phi$ is isomorphic, and indeed, by counting argument, it suffices to show it is injective or surjective.

This is where I am stuck. To show that the kernel of $\phi$ is trivial, it would suffice to prove the triviality of $\bigcap_{x\in S_n\setminus G} xGx^{-1}$, but I do not know why this is true. I do not have any idea how to show the surjectivity of $\phi$, either. I have not yet used specific properties of symmetric groups, so I believe I need to use them here.

I would be grateful if you could provide a clue.

We can suppose that $n>4$, to avoid trivialities. You can easily do the remaining cases simply by inspection.

$S_n$ acts transitively on the set of left cosets $S_n/G$ by multiplication on the left. This gives a morphism $S_n\to S(S_n/G)$. The kernel of the action is $K=\bigcap_{h\in S_n}hGh^{-1}$, which is a normal subgroup of $S_n$ contained in $G$. Since $S_n$ has exactly one non-trivial proper normal subgroup, which is of index $2$, we see that $K$ is trivial. The restriction to the action to $G$ is therefore also injective. $G$ acting on $S_n/G$ has one orbit of size $1$, namely $\{G\}$, so dropping that gives us an injective action from $G$ on a set of order $n-1$.