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$\DeclareMathOperator\Aut{Aut}\DeclareMathOperator\Inn{Inn}$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 ?

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    $\begingroup$ The automorphism group of $\mathbf{Z}\times\mathbf{Z}/2\mathbf{Z}$ is not cyclic of order $2$. Indeed, $w\mapsto -w$ is an automorphism of order $2$, and $(x,y)\mapsto (x,y+\pi(x))$ is another one, where $\pi$ is the projection modulo 2. Same remark for $\mathbf{Z}/4\mathbf{Z}\times\mathbf{Z}/2\mathbf{Z}$. $\endgroup$
    – YCor
    Oct 4, 2021 at 8:34
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    $\begingroup$ @Shika-Blyat indeed I was confused by your formulation. Is it a question about choice or not (I initially understood you were asking about f.g. abelian groups under ZF)? Maybe it would be useful to expand Karagila's assertion. Also see mathoverflow.net/a/198339/14094 which provides an infinitely generated abelian group with automorphism group of order 2. $\endgroup$
    – YCor
    Oct 4, 2021 at 8:45
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    $\begingroup$ I'm still confused about the question here - doesn't the answer by Asaf linked in the OP already answer the question negatively (by taking $\mathbb F=\mathbb F_3$)? $\endgroup$
    – Wojowu
    Oct 4, 2021 at 9:24
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    $\begingroup$ Well, the answer I posted only dealt with the choiceless case, so it's definitely not there. (Also, it's nice to see people read it, that answer became the basis for my masters thesis.) $\endgroup$
    – Asaf Karagila
    Oct 4, 2021 at 10:03
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    $\begingroup$ See Lemma 2 on second to last page, specifically property (c) of the group. $\endgroup$
    – Wojowu
    Oct 4, 2021 at 11:12

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