Fix a finite group $G$, and let $f_G:\mathbb{N}\rightarrow\mathbb{N}$ be defined by setting $f_G(n)$ to be the largest $m$ such that $S_n$ contains $m$ disjoint (pairwise-intersecting at 1) copies of $G$. In other words, $f_G(n)$ is the optimal number of $G$'s that can be packed into $S_n$ without unnecessary overlap. Clearly $f_G$ starts with a string of zeros, monotonically increases and is unbounded. It would seem spectacularly ambitious to actually compute this function for $G$ remotely complicated.

Q1 - Does $f_G$ determine $G$?

In case the affirmative does hold generally, I'd also be curious to know if a stronger result is true:

Q2 - Does there necessarily exist an $n$ (depending on $G$) such that $f_G(n)\neq f_{H}(n)$ for all $H$ distinct from $G$? In other words, can we be assured that a finite amount of computation would be sufficient to distinguish $G$ from all other finite groups?

An affirmative answer to Q2 would mean that $G$ is classified by a pair of (hopefully computable) integers.

Both answers are affirmative for groups of (very) small order since they're the only ones that even fit in the first few symmetric groups, but I don't see any hope to prove/disprove them in general.