[Note: This question is closed. It's current content reflects a draft of a potential new question, modified from the original by adding conditions to the premises; see comments]

Let $W,X$ be cancellative invertible-free [1] monoids. A map $e\colon W\rightarrow X$ is a *homography* [2] if it is non-decreasing in the prefix order and $e(1)=1$. The *shift* of a given homography $e$ by an element $w\in W$ is $d=e^w$ satisfying $e(w \,w') = e(w) \,d(w')$. (Note this is well-defined and $d(1)=1$ by cancellation property, and it exists because e is prefix non-decreasing.) Then $W$ acts on $D=e^W=\{e^w\colon w\in W\}$ on the right by shift, i.e. $d \cdot w = d^w$ for $d\in D$ and $w\in W$ is a right action, and $W$ acts on the set of homographies from $W$ to $X$ by shift as well.

Is the following a theorem?

Let $A$ be a set, $W$ a cancellative invertible-free monoid, and $\cdot\colon A \times W \rightarrow A$ a cyclic right $W$-action generated by an element $a_0\in A$, in symbols $A=\langle a_0\rangle=\{a_0\cdot w\colon w\in W\}$. Then there exists a cancellative invertible-free monoid $X$, set $B$ of homographies from $W$ to $X$, and bijection $\varphi \colon A \leftrightarrow B$ such that

$B$ is closed under shift, in symbols $B^w \subseteq B$ for all $w\in W$, and

when $W$ acts on $B$ by shift, $\varphi$ is an isomorphism of $W$-actions ($\varphi(a\cdot w) = \varphi(a)^w$ for all $a\in A$ and $w\in W$).

To convert this to a conjecture in operator theory, let $W$, $X$, and $e$ be as above, $R$ a commutative ring, and $C=R[e^W]$ the $R$-module of formal linear combinations of elements of $e^W$ with coefficients in $R$. Then for all $w\in W$, precomposition shift by $w$ is an $R$-linear operator on $C$ (when elements of $C$ are considered to be maps $c\colon e^W \rightarrow R$, this is $T_{w}\colon C\rightarrow C$ where $T_{w}(c)(d)=c(d^w)$).

Is the following a theorem?

Let $R$ be a commutative ring, $W$ a cancellative invertible-free monoid, $F$ an $R$-module, and $\cdot\colon F\times W\rightarrow F$ a cyclic $R$-linear $W$-action linearly generated by $f_0\in F$, in symbols $R\,f_0\cdot W = \{r\,f_0\cdot w\colon r\in R \wedge w\in W\} = F$ and $(r\,f + s\,f')\cdot w = r\,(f\cdot w) + s\,(f'\cdot w)$ for all $r,s\in R$, $f,f'\in F$, and $w\in W$. Then there exists a cancellative invertible-free monoid $X$, set $G$ of homographies from $W$ to $X$, and bijective $R$-linear map $U\colon F\leftrightarrow R[G]$ such that $U(f\cdot w) = T_w(U(f))$ for all $f\in F$ and $w\in W$.

[1] Can every cancellative invertible-free monoid be embedded in a group?

[2] Let a *line through the origin* of $W$ be a subset that contains the identity and is totally ordered by prefix, and let the *projective space* $P(W)$ be the set of lines through the origin of $W$. This differs from ordinary usage in that homography refers to a map of the argument of projective space here, whereas a it might have been defined as a map from $P(W)$ to $P(X)$ induced by an identity-preserving prefix non-decreasing bijection $e\colon W\leftrightarrow X$ of cancellative invertible-free monoids, following the Wikipedia article on homography. As defined here, a homography preserves lines through the origin.

1more comment