= Solution
\b[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 $f(1)=0\ne1$. 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>.
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