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.