Is any group the automorphism group of some (commutative) monoid?

Let $G$ be an arbitrary group. We construct an edge-colured graph $(V,E,C,\Gamma)$ having $G$ as automorphism group: As set of vertices, we take $V=G$, as set of edges, we take $E=\{\,g\to h: g,h\in G, g\ne h\,\}$ (i.e., we are dealing with thge complete directed graph on $G$). As set of colours, we take $C=G\setminus\{1\}$, and let the colouring $\Gamma\colon E\to C$ be given as $\Gamma(g\to h)=hg^{-1}$.

Claim. Let $G$ ba any group and $(V,E,C,\Gamma)$ as described above. Then $\operatorname{Aut}(V,E,C,\Gamma)\cong G$.

Proof. If $a\in G$, then $\phi_a\colon V\to V$, $g\mapsto ga$ induces an automorphism. Indeed, if $g\to h$ is an edge then so is $ga\to ha$, and we have $\Gamma(ga\to ha)=ha(ga)^{-1}=hg^{-1}=\Gamma(g\to h)$. As $\phi_a(1)=a$, we see that $a\mapsto \phi_a$ is a monomorphism $G\to \operatorname{Aut}(V,E,C,\Gamma)$. On the other hand, let $\phi$ be an arbitrary automorphism of $(V,E,C,\Gamma)$ and let $a=\phi(1)$. Then $\phi=\phi_a$ because for any $g\ne 1$, the edge $a\to \phi(g)$ must have the same colour as $1\to g$, hence $\phi(g)a^{-1}=g$, $\phi(g)=ga$. $\square$

Let $(V,E,C,\Gamma)$ be an edge-coloured directed graph, where wlog. $C$ is an ordinal. We define an (un-coloured) directed graph $(V',E')$ as follows. Let $V'$ consist of

• a vertex $v$ for each $v\in V$
• a new vertex $\hat v$ for each $v\in V$
• a new vertex $(\beta,v, w)$ for each edge $v\to w$ in $E$ and $\beta\le \Gamma(v\to w)$

Let $E'$ consist of

• an edge $v\to \hat v$ for each $v\in V$
• an edge $v\to (0,v,w)$ for each edge $v\to w$ in $E$
• an edge $(\Gamma(v\to w),v,w)\to w$ for each edge $v\to w$ in $E$
• an edge $(\beta,v,w)\to (\gamma,v,w)$ whenever $0\le \beta<\gamma\le \Gamma(v\to w)$

Claim. Let $(V',E')$ be obtained as described from $(V,E,C,\Gamma)$. we have $\operatorname{Aut}(V'E')\cong\operatorname{Aut}(V,E,C,\Gamma)$.

Proof. Given an automorphism $\phi$ of $(V,E,C,\Gamma)$, we can readily find a corresponding automorphism of $(V',E')$ by mapping vertices $$\begin{align}v&\mapsto \phi(v),\\ \hat v&\mapsto \widehat{\phi(v)},\\ (\beta,v,w)&\mapsto (\beta,\phi(v),\phi(w)).\end{align}$$ This gives us a monomorphism $\operatorname{Aut}(V,E,C,\Gamma)\to \operatorname{Aut}(V'E')$. We need to show that it is onto. Vertices of the second type ("$\hat v$") are characterized by the fact that they have no successors. Vertices of the first type ("$v$") are characterized by the fact that they have a successor of the first type. Thus also vertices of the third type are characterized. An automorphism $\phi$ of $(V',E')$ must respect types. Moreover, vertices of the third type with ordinal $0$ are characterized by being of third type and having a type one predecessor, so this must also be respected by an automorphism. Assume $\phi((0,v,w))=(0,a,b)$ where $v\to w$ is an edge in $E$ of colur $c$. Then $\phi(v)=a$ by looking at the unique predecessors and $\phi(\hat v)=\hat a$ by their type 1 successors. I claim that $\phi((\beta,v,w))=(\beta,a,b)$ for all $\beta\le c$. Indeed, if $\beta>0$ and the claim holds for all smaller ordinals, then $\phi((\beta,v,w))$ is determined among all type 3 vertices by the fact that precisely all $(\gamma,a,b)$ with $\gamma<\beta$ are predecessors. After that, note that $w$ is the unique successor of $(c,v,w)$, hence $(c,a,b)$ must have a sucessor of second type; this must be $b$ and therefore $\Gamma(a\to b)=c$ and $\phi(w)=b$. We conclude that $\phi$ induces a permutation of $V$ and is uniquely determined by that permutation, and if $v\to w$ is an edge in $E$ then $\phi(v)\to \phi(w)$ is also an edge and is of the same colour. In other words, our monomorphism $\operatorname{Aut}(V,E,C,\Gamma)\to \operatorname{Aut}(V'E')$ is in fact an isomorphism. $\square$

Next, $(V,E)$ be a directed graph. We define an undirected graph $(V',E')$ as follows.

Let $V'=V\cup (E\times\{1,2,3,4\}$, i.e., $V'$ consists of

• a vertex $v$ for each $v\in V$
• new vertices $(v,w)_1, (v,w)_2, (v,w)_3, (v,w)_4$ for each edge $v\to w\in E$

and let $E'$ consist of

• edges $v\leftrightarrow (v,w)_1$,
• $v\leftrightarrow (v,w)_2$,
• $v\leftrightarrow (v,w)_3$,
• $(v,w)_1\leftrightarrow (v,w)_2$,
• $(v,w)_2\leftrightarrow (v,w)_3$,
• $(v,w)_2\leftrightarrow (v,w)_4$,
• $(v,w)_3\leftrightarrow w$ for each edge $v\to w$ in E$

Claim. With $(V',E')$ obtained from $(V,E)$, we have $\operatorname{Aut}(V,E)\cong \operatorname{Aut}(V',E')$.

Proof. As before, from an automorphism $\phi$ of $(V,E)$, we readily obtain an automoprhism of $(V','E')$ that maps $v\mapsto \phi(v)$ and $(v,w)_i\mapsto (\phi(v),\phi(w))_i$., thereby obtaining a monomoprhism $\operatorname{Aut}(V,E)\to\operatorname{Aut}(V',E')$.

Now let $\phi$ be any automophism of $(V',E')$. Observe that vertices of the form $(v,w)_1$ are precisely those that are of degree $2$ and with an edge between their neighbours. These two neighbours have precisely one other common neighbour, namely the corrsponding $(v,w)_3$. Exactly one of these common neighbours, namely $(v,w)_2$, has a neigbour of degree $1$, which is $(v,w)_4$; the other common neighbour must be $v$. And the only neighbour of $(v,w)_3$ not yet mentioned is $w$. We conclude that $\phi$ must map $(v,w)_1$ to some $(a,b)_1$ and that then $\phi(v)=a$, $\phi(w)=b$, $\phi((v,w)_i)=(a,b)_i$. In particular $\phi$ induces a permutation of $V\subset V'$ and such that $(V,E)$ has an edge $\phi(v)\to\phi(w)$ whenever $v\to w$ is an edge. In other words, our monomorphism $\operatorname{Aut}(V,E)\to\operatorname{Aut}(V',E')$ is an isomorphism. $\square$

Finally, given an undirected graph $(V,E)$, we construct a commutative semigroup with the same automorphism group: Let $S=V\cup E\cup\{0\}$ and declare

• $vv=v$ if $v\in V$
• $ve=ev=e$ if $v\in V$ is an end vertex of edge $e\in E$
• $xy=0$ otherwise.

Any automorphism of the graph gives rise to an automorphism of $S$. On the other hand, any automorphism of $S$ must respect vertices (idempotent elements) and edges (other non-zero elements)products) and vertex-edge incidence (vertex and edge with non-zero product)

(This essentially repeats the construction given in the accepted answer to one of the referenced similar questions).

This shows the final

Claim. Let $(V,E)$ be an undirected graph. Then there exists a commutative monoid $S$ such that $\operatorname{Aut}(V,E)\cong\operatorname{Aut}(S)$. $\square$

Combining all of the above, we see that for every (finite or inifnite) group $G$, there exists a commutative monoid $S$ such that $$\operatorname{Aut}(S)\cong G. $$