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 $c$.