 Solution

It can be false even on a two-element finite set. On $X=\{0,1\}$, let $f$ and $g$ both be the constant map with value $0$. Then $fg=g$, but $f(1)=0\ne 1$. In general the equation says exactly that $f$ fixes the image $g(X)$; it says nothing about other points. If $g$ happens to be a surjection, it does force $f$ to be the identity map, but finiteness alone does not ensure this. The empty set and a singleton have only one self-map, so the assertion is trivially true on those sets.