How to prove $|\text{Aut(G)}|=2\implies |G|=3$

Question

As suggested in the title I'm trying to prove $|G|=3$ given that $|\text{Aut}(G)|=2.$

First I thought that: for an automorphism, we have no choice for the identity element because the identity must go to the identity and for the other non-identity elements there are exactly $2$ permutations. So if $|G|=n$ then $(n-1)!=2\implies n=3$.

But I realized that the orders of elements are being preserved by automorphisms. So any permutation is not allowed for considering the automorphisms.

Please help me to get the idea of the problem. Thank you.

Answer 1

What you are trying to prove is incorrect, since each of the cyclic groups $C_3,C_4,$ and $C_6$ has automorphism group containing exactly $2$ elements. So it cannot be the case that $|\textrm{Aut}(G)|=2$ implies $|G|=3$.