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The object $(\mathbb{Z}, +)$ is interesting. It intuitively has lots of internal symmetry, but $\text{Aut}(\mathbb{Z}, +)$ is only $S_2$. Our map is completely determined by where we send $1$ and we have to send it to $\pm 1$.

$(\mathbb{Z}, +)$, however, is rich in endomorphisms.

I had a silly idea to look at the automorphism group of the endomorphism monoid after reading this MO post on automorphism towers from a long time ago. This post has nothing to do with the meat of that post, I'm just borrowing the idea of stacking $\text{Aut}$-like operations.

Do we know the automorphism group of the endomorphism monoid of Abelian groups in general?

Consider $(\mathbb{Z}, +)$. An endomorphism of $(\mathbb{Z}, +)$ is completely determined by where it sends $1$ and looks like multiplication by a fixed integer so we get $(\mathbb{Z}, *)$. The symmetries of $(\mathbb{Z}, *)$, though, are the bijections that fix the primes setwise (and 0 and 1 elementwise). There are infinitely many primes, so $\text{Aut}(\text{End}(\mathbb{Z}, +))$ is $S_\omega$, the symmetric group on $\omega$.

Consider $(\mathbb{Z}/p\mathbb{Z}, +)$ where $p$ is prime. An endomorphism is again completely determined by where it sends $1$ and looks like mulitplication by a fixed element, so we get $(\mathbb{Z}/p\mathbb{Z}, *)$. However, the multiplicative group of a finite field is cyclic, so we get $C_{(p-1)}$. The multiplicative monoid of a field has the same symmetries as its group of units, obtained by discarding zero. By a result quoted in this hint in an answer, that means that $\text{Aut}(\text{End}(\mathbb{Z}/p\mathbb{Z}, +))$ is isomoprhic to $\mathbb{Z}^*_{p-1}$.

For $(\mathbb{Z}/4\mathbb{Z}, +)$, the endomorphisms are multiplication by 0, 1, 2, or 3. They have no nontrivial symmetries. 0 and 1 must be fixed, as must 2. So $\text{Aut}(\text{End}(\mathbb{Z}/4\mathbb{Z}, +))$ is $S_1$.

For $(Z_2 \times Z_2, +)$, the endomorphisms are 2x2 matrices of the form $ \begin{bmatrix} a & b \\ c & d \end{bmatrix}$. The transpose is injective, but reverses composition so we can't use it. However, conjugation by $\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$ does not reverse composition, so $S_2$ is certainly a subgroup of $\text{Aut}(\text{End}(Z_2 \times Z_2, +))$, although I'm not sure the true identity of the group.

Shaun
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Greg Nisbet
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    Why ignore the addition on the endomorphism ring? If the goal is just to get an interesting monoid, why restrict your attention to endomorphism monoids? – Qiaochu Yuan Dec 31 '22 at 05:10
  • @QiaochuYuan I didn't realize until after writing the question that the endomorphisms of an Abelian group are typically organized into a ring. Yes, the goal is just to get an interesting monoid. I assumed that asking "what do we know about possible symmetries of monoids?" was way too broad. – Greg Nisbet Dec 31 '22 at 05:30
  • We should be able to prove that every group arises as the automorphism group of a monoid. There are various results of this form that imply each other, e.g. every group is the automorphism group of a graph and there are ways to turn a graph into a monoid with the same automorphism group. – Qiaochu Yuan Dec 31 '22 at 08:48

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