$\DeclareMathOperator\Aut{Aut}\DeclareMathOperator\Inn{Inn}$First I hope my question belongs here, please let me know if it doesn't.
It isn't too hard to show there is no groups $G$ such that $\Aut(G) \simeq \mathbb{Z}/p\mathbb{Z}$ with $p \geq 3$ prime. My question is: what happens when $p = 2$ ?
First, using the fact that if $\Inn(G)$ is finite and cyclic then $G$ is abelian, we get that $G$ is abelian and $\Inn(G)=\{\mathrm{Id}\}$.
Assuming $G$ is finitely generated, we can write $G \simeq \mathbb{Z}^n \times \mathbb{Z}/p_1\mathbb{Z} \times \cdots \times \mathbb{Z}/p_k\mathbb{Z}$.
Then, if $n \geq 2$, we get at least $4$ automorphisms.
If any $p_i$ is distinct of $2,3,4,6$, we would have $$2 < \varphi(n) =|\Aut(\mathbb{Z}/p_i\mathbb{Z})| = |(\mathbb{Z}/p_i\mathbb{Z})^\times| \leq |\Aut(G)|.$$
If we have any factor of the form $(\mathbb{Z}/2\mathbb{Z})^2, (\mathbb{Z}/3\mathbb{Z})^2,(\mathbb{Z}/4\mathbb{Z})^2, (\mathbb{Z}/6\mathbb{Z})^2$, then we can exhibit $3$ atleast automorphisms.
After checking the remaining products by hand, we get that the followings satisfy $\Aut(G) \simeq \mathbb{Z}/2\mathbb{Z}$:
$$\mathbb{Z}, \mathbb{Z}/3\mathbb{Z}, \mathbb{Z}/4\mathbb{Z}, \mathbb{Z}/6\mathbb{Z} \simeq \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$$
We can also prove the second automorphism must be $x \mapsto -x$, assuming AoC:
If $x = -x$, then $G$ is a vector space over $\mathbb{Z}/2\mathbb{Z}$.
When $\dim(G) = 0,1$, there is no nontrivial automorphism, as for when $\dim(G) > 2$, permutations of elements of a basis gives atleast $\dim(G)! > 2$ automorphisms.
When $\dim(G) = 2$, we can simply exhibit $3$ automorphisms.
Without any form of choice this doesn't hold though, as shown here by Asaf Karagila, leading me to the question: does the statement $\Aut(G) \simeq \mathbb{Z}/2\mathbb{Z}$ iff $G$ is one of the finitely generated groups I gave above, depend on choice somehow ?
