The number of involutions in a permutation group If $G$ is a group let $I(G)$ be the number of involutions (elements of order 2) in $G$. My question is then easily stated: does there exists a constant $C > 1$ such that for every $n \ge 1$ and every subgroup $G \subset S_n$ of the symmetric group $S_n$ we have 
$$
C^{-n} |G|^{\frac 1 2} \le I(G)+ 1 \le C^n |G|^{\frac 1 2}. 
$$
Some remarks : 
*This is true for primitive groups, since those groups are either very small (of size at most $D^n$ for some absolute $D$) or they have to be the full symmetric or alternating groups (for which the number of involutions is precisely known). 
*Using iterated wreath products decompositions and the previous remark it is possible to show something like 
$$
C^{-n\log\log(n)} |G|^{\frac 1 2} \le I(G)+ 1 \le C^{n\log\log(n)} |G|^{\frac 1 2}. 
$$
(This is actually good enough for the application I have in mind but I was wondering whether a sharper result to which I could refer existed). 
*The lower bound would be sharp, since for example a 3-Sylow of $S_n$ is of size roughly $3^{n/2}$ as $n$ goes to infinity and contains no involutions. 
*This question: Number of involutions in a finite group seems like it could be relevant but estimating the number of conjugacy classes in this setting seems to have to be rather involved. 
Edited to add :
*I am interested only in the exponential aspect of the bound, but one might also ask for optimal $c < 1 < C$ such that 
$$
c^n |G|^{\frac 1 2} \le I(G)+ 1 \le C^n |G|^{\frac 1 2}. 
$$
(see Yves' comments below). 
*With this notation the third point above (which I edited for clarity) gives an upper bound $< 1$ for $c$; looking at subgroups of exponent 2 in $S_n$ (for example $(\mathbb Z/2\mathbb Z)^{n/2}$) also gives a lower bound $> 1$ for $C$. An upper bound for $C$ is given by Geoff Robinson's answer, now we only lack a lower bound $>0$ for $c$. 
 A: I can answer in one direction. L.G. Kovacs and I proved ( around 1993) that any finite subgroup $G$ of $S_{n}$ has at most $5^{n-1}$ conjugacy classes. Later authors showed that it is possible to replace $5$ by a smaller constant, but the existence of such a constant seems to be enough for this question. It follows that any  subgroup $G$ of $S_{n}$ has at most $\sqrt{5}^{n-1}|G|^{\frac{1}{2}} -1$ involutions, using properties of the Frobenius-Schur indicator, as in the question you refer to.
Later edit: It might be worth recasting the problem ( this is not so relevant for the direction already proved above): if the group $G$ above has no involutions, ( equivalently, has odd order), then $G$ is certainly solvable, so we have $|G| \leq 24^{\frac{n-1}{3}}$ by a result of J.D. Dixon ( again, the constant can be improved here, but its existence is enough for present purposes). This shows that $c = 24^{\frac{1-n}{6}}$ works above when $|G|$ is odd, and clearly $C = 1$ works here too ( for the other direction).
So we only now need to consider the case when $G$ has even order. It is well known that every $2$-subgroup of $S_{n}$ has order at most $2^{n-1}$, so that a Sylow $2$-subgroup of $G$ has order at most $2^{n-1}$ and, in particular, $G$ certainly has at most $2^{n-1}$ conjugacy classes of involutions.
If there is a positive constant $c$ as asked for in the later version of the question, then $G$ has an involution $t$ such that $[G:C_{G}(t)] > \left(\frac{c}{2} \right)^{n}|G|^{\frac{1}{2}}$ and hence $|C_{G}(t)| < \left(\frac{2}{c} \right)^{n}|G|^{\frac{1}{2}}.$ 
It follows that the question is equivalent to asking whether it is true that there are positive (finite) constants $d,D$ such that whenever $G$ is an even order 
subgroup of $S_{n},$ there are involutions $t,u \in G$ such that 
$|C_{G}(t)| \leq d^{n}|G|^{\frac{1}{2}}$ and $|C_{G}(u)| \geq D^{n}|G|^{\frac{1}{2}}.$
