01-28 Update: In the first version I was claiming that the authors were not explicitly or implicitly but I was wrong so I change my question [long explaination at the end of the question]
Two elements of a non-commutative Monoid $a,b\in M$ are conjugate if exists an element $\varphi\in M$ such that the conjugacy equation holds
$$\varphi a=b\varphi$$
I will call $\varphi$ "a solution" of the congugacy equation and denote the set of solutions with $\mathcal C(a,b)$ since I don't know the common notation.
$$\mathcal C(a,b):=\{\varphi:\varphi a=b\varphi\}$$
CE:=conjugacy equation
I'm reading a paper on Meromorphic solution of conjugacy equations that talks about the connection between $\mathcal C(a,b)$ and $\mathcal C(a,a)$, the solutions of a special case kind of conjugacy equation (where $a=b$) that is called "permutable functional equation" (PFE).
The very first lemma they give is the following
LEMMA: let $\varphi_0$ be a particular solution of [the C.E.] Then every solution $\varphi$ of [C.E.] is given by $$\varphi=\varphi_0\alpha$$ where $\alpha$ is a solution* of [PFE].
(*)It assumes that $\alpha\in \mathcal C(a,a)$ And then they prove it as follows
PROOF: since $\varphi$ is a solution of [CE], we have $\varphi_0 a=b\varphi_0$. For any solution $\alpha$ of [PFE], let $\varphi=\varphi_0\alpha$ , then
$$\varphi a=(\varphi_0\alpha) a=\varphi_0a\alpha=b(\varphi_0\alpha)=b\varphi$$ This completes the proof.
I can't understand how this proof is enough to prove the lemma. It seems that everything it does is to prove that
if $\varphi_0\in \mathcal C(a,b)$ and $\alpha \in \mathcal C(a,a)$ then $\varphi_0\alpha \in\mathcal C(a,b)$
In other words $\mathcal C(a,b)$ is closed under right multiplication by elements of $\mathcal C(a,a)$: $\forall\varphi_0\in\mathcal C(a,b) $
$$\varphi_0\mathcal C(a,a)\subseteq \mathcal C(a,b)$$
but the lemma seems to claim something stronger (in my opinion): fixed a $\varphi_0\in \mathcal C(a,b)$
For all $\varphi \in \mathcal C(a,b)$ exists an $\alpha \in C(a,a)$ such that $\varphi=\varphi_0\alpha$
that to me seems more like claiming that $\varphi_0\mathcal C(a,a)= \mathcal C(a,b)$ for every $\varphi_0$.
If $M$ is a group then $\alpha$ is given by $\varphi_0'\varphi$ that in fact satisfies the PFE of $\alpha$
$\varphi_0'\varphi a=\varphi_0' b\varphi=a\varphi_0'\varphi$ thus $\varphi_0'\varphi\in \mathcal C (a,a)$
So $\varphi\mapsto \varphi_0'\varphi$ defines a bijection $\mathcal C (a,b)\rightarrow \mathcal C (a,a)$ but the paper doesn't assumes invertibility since it is talking about composition of functions(**) and in the proof uses only the associativity axiom.
I feel like I'm missing something, if it is too trivial I'd like hints too.
Thanks in advance and sorry for my bad english.
(**) Note about the paper: The Lemma 1 (pag 2 of the link) says that $\varphi_0$ is a solution of the CE (referred as Eq. (1) in the paper) but the authors seems to cosider only the homeomorphic solutions $\varphi$ of the equation $\varphi\circ f=g\circ\varphi$ thus $\varphi_0$ is invertible and should complete the proof.
I just read the reference paper and you forgot an important hypothesis. In the paper, $\varphi$ denotes an homeomorphism, which means in terms of monoids, that $\varphi$ and $\varphi_0$ are inversible.
The conclusion is now easy. Suppose that $\varphi$ and $\varphi_0$ are in $C(a,b)$ and let $\alpha = \varphi_0^{-1}\varphi$. Then $\varphi = \varphi_0\alpha$ $\varphi a = b\varphi$ and $\varphi_0 a = b\varphi_0$. It follows that $\varphi_0\alpha a = b \varphi_0\alpha = \varphi_0 a\alpha$ and finally $\varphi_0^{-1}\varphi_0\alpha a = \varphi_0^{-1}\varphi_0 a\alpha$, that is $\alpha a = a \alpha$. Thus $\alpha \in C(a,a)$.