TITLE: Let $G$ be a group of order $n$ where every non identity element is of order $2$. Then, ${\rm Aut}(G) $ is isomorphic to $S_{n-1}$
QUESTION [5 upvotes]: Let $G$ be a group of order $n$ where every non identity element is of order $2$. Then,  ${\rm Aut}(G) $ is isomorphic to $S_{n-1}$. Is this true?
I read that ${\rm Aut}(K_4)$ is isomorphic to $S_3$. We can see this because of the permutations of the non identity elements, all of which have order $2$. So for each non identity element, we have $3$ choices for it's image. Proceeding recursively, we get $3!$ and  ${\rm Aut}(K_4)$  is isomorphic to $S_3$.
Can we further generalize this idea to groups in which all non identity elements are of order $2$?
Thanks in advance!

REPLY [4 votes]: Let $a$ and $b$ be two distinct non-identity elements of $G$. If $\tau$ is an automorphism of $G$ with $\tau(a)=a$ and $\tau(b)=b$, then we must have $\tau(ab)=ab$.
If $n>4$, then there exists an element $c \in G \setminus \{e,a,b,ab\}$. In that case there is no automorphism of $G$ with $\tau(a)=a$, $\tau(b)=b$, and $\tau(ab)=c$, so not all permutations of $G \setminus \{e\}$ are possible, and hence $|{\rm Aut}(G)| < |S_{n-1}|$.