$\operatorname{Aut}(V)$ is isomorphic to $S_3$

abstract-algebra group-theory finite-groups

This is my answer. Additional comments are always welcome.

Let $f:V\rightarrow V$ be a permutation that fixes the identity element. By definition, this map is bijective. $V$ has the property that the product of any 2 distinct non-identity elements is the 3rd non-identity element. This implies $f(vv^\prime)=f(v)\cdot f(v^\prime)$ for all $v,v^\prime \in V$ with $v,v^\prime\neq e$. For products involving the identity element, $f(ev)=f(e)\cdot f(v)=e\cdot f(v)=f(v)$ for all $v\in V$. Also, since all non-identity elements of $V$ have order 2, $f(v^2)=f(vv)=f(v)\cdot f(v)=e$ for all $v\in V$. Hence, any permutation of the elements of $V$ that fixes the identity element is an automorphism of $V$. Since these permutations exhaust all possible automorphisms of $V$, they are the elements of Aut($V$).

Since the $3!=6$ elements of Aut($V$) represent all permutations of 3 objects, Aut($V$) is isomorphic to S$_3$.

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