Show that there is no subgroup of $S_n$ of order $(n-1)!/n$.

This conjecture is true, here is a sketch of a proof.

You are looking for a group $G$ of index $n^2$ in $S_n$.

$G$ is either

(1) intransitive,

(2) transitive but imprimitive,

(3) primitive.

In the first case, we must have $G\leq S_a \times S_{n-a}$ for some $1\leq a\leq n/2$. But this implies that ${n}\choose{a}$ divides $n^2$. It is not hard to see that this implies that $a=1$ hence $G$ is contained in $S_{n-1}$. We are now asking about a subgroup of index $n$ in $S_{n-1}$, and we can repeat the whole argument here to find that the only option is $F_{20}$ in $S_5$.

Suppose now that $G$ is transitive but imprimitive, so $G\leq S_{n/a}^{a} \rtimes S_a$, for some divisor $1<a<n$ of $n$. As before, this implies that $\frac{n!}{a!(n/a)!^a}$ divides $n^2$ which, with a bit of work, can be shown not to happen. (The left side is typically much bigger than $n^2$.)

Finally, if $G$ is primitive, then it is quite small, in fact $|G|\leq 4^n$, for example by

Praeger, Cheryl E.; Saxl, Jan On the orders of primitive permutation groups. Bull. London Math. Soc. 12 (1980), no. 4, 303–307.

and, as in the imprimitive case, this is too small to have index $n^2$. (Except for small $n$ which need to be checked by hand.)