Let $G$ be a finitely generated group. I am trying to count the number of permutation representations on $n$ elements, i.e. homomorphisms from $G$ to the symmetric group $S_n$. Equivalently this is the number of ways that $G$ can act on the set $\{1,2,\ldots,n\}$. Let $a_{G,n}$ be the number of homomorphisms from $G$ to $S_n$. Let $\alpha(G) = \lim_n \frac{\log(a_{G,n})}{n\log n}$. How can we determine $\alpha(G)$?

For example, if $G = \mathbb Z$ we have $a_{G,n} = n!$, since a homomorphism $\mathbb Z \to S_n$ is determined by the image of its generator. By Stirling's approximation we have $\alpha(\mathbb Z) = 1$. More generally, for $G = F_s$ the free group on $s$ generators, we get $\alpha(F_s) = s$. For $G = \mathbb Z/2$, a morphism $G\to S_n$ can be seen as an order 2 element in $S_n$. We can create such a permutation by dividing the elements of $\{1,2,\ldots,n\}$ in to pairs, which can be done in $\frac{n!}{(n/2)!2^{n/2}}$ ways. This is asymptotically $\exp(\frac12 n\log n)$. There are some more permutations of order 2 that have fixed points, but not enough to change the asymptotic result. So $\alpha(\mathbb Z/2) = \frac12$. More generally, for any finite group $G$ we have $\alpha(G) = 1-\frac1{|G|}$.

How can we determine $\alpha(G)$ for an arbitrary group?

Enumerative Combinatorics, vol. 2, Exercise 5.13. $\endgroup$