in my notations, $S_n$ is the symmetric group and $A_{n+1}$ the alternating group.
I showed that if $n$ is even it is never possible because you need to have $n! | n! \cdot\dfrac{n+1}{2}$.
So we need $n$ odd.
I also showed that it is impossible for $n=3$ because $A_4$ has no subgroups of order 6.
Now, for other odd values of $n\ge 5$, we can use the fact that $A_{n+1}$ is simple, but how to go on from here ?
The links given as duplicate do not provide a clear answer and I am still stuck on this. Here is what I have :
If you consider transposition $(1\quad 2)$, the permutations that commute with this transposition are the ones that fix 1 and 2, and the ones that switch 1 and 2. So that the size of the centralizer of $(1\quad 2)$ is $2(n-2)!$.
Now, because $(1\quad 2)$ is of order 2, it's image must also be an element of order 2 in $A_{n+1}$, so that it must be the product of $2k$ disjoint transpositions. If we now consider the centralizer of the product of $2k$ disjoint transpositions in $A_{n+1}$, we find that a permutation in the centralizer must fix or switch each of the transpositions, leaving $2^{2k}$ choices. Then, we must either pick a permutation of $A_{n+1-2k}$ or $S_{n+1-2k}\setminus A_{n+1-2k}$ to make sure the parity is even. Leaving in total :
$$ (n+1-2k)!\cdot 2^{2k-1}$$
Permutations in the centralizer of the product of $2k$ transpositions.
If we had $(n+1-2k)!\cdot 2^{2k-1}< 2(n-1)!$ we would be done, but it is not always true.
Best Answer
If we had $S_n$ isomorphic to a subgroup of $A_{n+1}$, then it would have index $\frac{n+1}{2}$, so the action of $A_{n+1}$ on the cosets of this subgroup gives a nontrivial group homomorphism $A_{n+1}\rightarrow S_{\frac{n+1}{2}}$, which is not possible for $n\geq 4$, since $|A_{n+1}|>|S_{\frac{n+1}{2}}|$, and $A_{n+1}$ is simple.