Let $X$ be a set and let $(X^X,\circ)$ denote the monoid of all maps $f: X\to X$, together with composition. Let $(\text{Sym}(X),\circ)$ be the group of all bijections from $X$ to itself.


Does there exist a monoid homomorphism $h:X^X \to \text{Sym}(X)$ such that for every group $G$ and every monoid homomorphism $f: X^X\to G$ there is a homomorphism $f': \text{Sym}(X)\to G$ such that $f = f'\circ h$?



Do monoid homomorphisms from $X^X$ to a group factor through $\text{Sym}(X)$?
Yes for trivial reasons. Let $c$ be a constant map. Then for any two $f$ in $X^X$ we have $ c \circ f = c$. Hence the image of $f$ under any homomorphism to a group must be trivial.


