Good, Anindya! Right, saying that a map from a monoid to itself \$$F: X \to X\$$ preserves multiplication quickly implies that \$$F(1)\$$ is an idempotent that commute with everything _in the range_ of \$$F\$$, but no more... so on the principle that "you don't get something for nothing", there should be examples where \$$F(1)\$$ doesn't commute with everything in \$$X\$$. Then the challenge is to find a counterexample... a challenge you met.