Let $M$ be a monoid that acts transitively from the right on a finite set $X$.
Assume furthermore that the action of $M$ on $X$ induces for every $m \in M$ a bijection on $X \to X, x \mapsto x.m$.
Let
$$M\_X := \{ m \in M \; | \; \forall x \in X : x.m = x \}$$
be the fixer of $X$.


Question 1: Is there always a finite subset $S$ of $M$ such that $S M\_X = M$?


(here $S M\_X$ denotes the set $\{ sm \; | \; s \in S, m \in M\_X \}$)


Question 2: Assume that there is such a set $S$ as in question 1. Given elements $s \in S$ and $m \in M$ is there always an element $s' \in S$ such that $ms' \in s M\_X$?


(here $s M\_X$ denotes the set $\{ s m \; | \; m \in M\_X \}$)


The answer to both questions is positive if either $M$ is a group and also if $M$
is the monoid of natural numbers with addition. In the case where $M$ is a group the affirmation in question 1 results from the fact that there are only finitely many cosets of $M\_X$ and question 2 from the fact that $M$ operates transitively on these cosets.



Cosets of the fixer of an action of a monoid on a finite set
The answer is no. Map the free monoid $M$ on $\{a,b\}$ onto $\mathbb Z/2$ by $a$ maps to $1$ and $b$ maps to $0$. Then $M$ acts transitively on the right of $\mathbb Z/2$ via bijections by applying the homomorphism and doing the regular representation. 


The fixer is all words with an even number of $a$'s. Consider the words $b^ka$ with $k>0$. No nonempty suffix is in the fixer so all these must belong to any potential $S$. 


