TITLE: Is the functor $\operatorname{Aut}(X): \operatorname{Mon} \longrightarrow \operatorname{Grp}^f$ representable ?
QUESTION [5 upvotes]: Let $X$ be a finite set. It defines a functor from the category of (finitely generated) commutative monoids to the category of groups as follows :
$$\operatorname{Aut}(X):M\mapsto \operatorname{Aut}_M(M\times X) $$
where $\operatorname{Aut}_M$ denotes the group of $M$-equivariant automorphisms of a set on which $M$ acts (clearly, $M$ acts on $M\times X$ by multiplication on the $M$ factor). Is this functor representable ?
Motivation
This is an attempt to better understand (finite) sets as (finite) vector spaces over "the field with one element" $\mathbb F_1$, since in the case of a field $k$, we have the corresponding functor for a given vector space $E$ from finitely generated commutative $k$-algebra to groups:
$$\operatorname{GL}(E):R\mapsto \operatorname{Aut}_R(R\otimes_k E)$$
which is represented by the algebra $$k[m_{11},\dots,m_{nn},t]/(t\cdot\operatorname{det}(M)-1)$$
where $\operatorname{det}(M)$ is the usual matrix determinant, which is a polynomial expression in the $m_{ij}$'s.
What I've tried so far
For $M$ a commutative f.g. monoid, there is a natural isomorphism $$\operatorname{Aut}_M(M\times X) \simeq \operatorname{Hom}(X,M^\times)\times \operatorname{Aut}(X)$$
where $M^\times$ is the group of invertible elements of $M$. This comes from the fact that if $\varphi$ is an $M$-automorphism of $M\times X$, then it is defined by its images on the pairs $(1,x)$ where $1$ is the neutral element of $M$, by $M$-equivariance, so there is a map $\mu:X \to M$ and a map $\sigma\in\operatorname{End}(X)$ such that $$\varphi(m,x)=(m\cdot\mu(x),\sigma(x)).$$
The fact that $\varphi$ is an isomorphism implies that $\mu$ has values in $M^\times$ and that $\sigma\in\operatorname{Aut}(X)$.
I know that $M\mapsto\operatorname{Hom}(X,M^\times)$ is representable (by the free abelian group generated by $X$). I am however stuck on how to deal with the $\operatorname{Aut}(X)$.

REPLY [4 votes]: The functor you describe is not representable. In the first place, a representable functor lands in $\mathbf{Set}$, not $\mathbf{Grp}$. But a more fundamental problem is that (in general) it does not preserve the terminal object: $\mathrm{Aut} (X) (1) \cong \mathrm{Aut} (X)$, and $\mathrm{Aut} (X) \cong 1$ if and only if $X \cong 1$.
Personally, I do not believe that the category of $\mathbb{F}_1$-modules is literally $\mathbf{Set}$ (or for that matter $\mathbf{Set}_*$). There are other structures that have the "correct" automorphism groups: for instance, for any set $X$, the power set $\mathscr{P} (X)$ is a join semilattice, and its automorphism group is canonically isomorphic to $\mathrm{Aut} (X)$, because $\mathscr{P} (X)$ is complete and atomic.