So I was considering the following question.
Is there a group of size n! (or less), that contains a larger number of subgroups than $$S_n$$? In thinking about this problem one obvious contender that came to mind is if we consider the largest power of 2 less than or equal to n! then we have the group $$ \Bbb{Z}_2 \times \Bbb{Z}_2 ... \Bbb{Z}_2 $$.
Naturally this has $$\ge 2^{\lfloor\log_2(n!)\rfloor}$$ different subgroups (from selecting some combination of crossing the subgroups $$\Bbb{Z}_2$$.
But do there exist ways to create even more subgroups than that? An example that came to mind was
$$ G_1 \times G_2... $$ whereas the total number of subgroups of this group obeys (let $O(G)$ denote the number of subgroups of G)
$$ O(G_1 \times G_2... \times G_n) \ge \times O(G_2) ... \times O(G_n)$$
So that was one layer of analysis, but here's another question, that follows up naturally.
Do there exist groups of size n with more unique subgroups than the symmetric group?
I think it is known that the number of subgroups of an elementary abelian $2$-group is larger than the number of subgroups of a symmetric group of roughly the same order.
For reference, the corresponding OEIS sequences are:
https://oeis.org/A006116
and
https://oeis.org/A005432
In particular, you can see that, already, the elementary abelian $2$-group of order 64 has 2825 subgroups while the one of order 128 has 29212. By comparison, the symmetric group of order 120 only has 156.
Asymptotically, the number of subgroups of the elementary abelian group of order $2^n$ grows like $c^{n^2}$ (from the OEIS page) while the number of subgroups of the symmetric group of order $n!$ grows at most like $d^{n^2}$. Since $n!$ is much larger than $2^n$, it should be an exercise to deduce the result from this.
(The constant $c$ is actually known, while bounds on $d$ are known, in case this is useful.)