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Find all endo-functions $f$ on a commutative semigroup $(\mathbb{S},*)$ such that $f(x*y) = f(f(x)*f(y))$.

Typical case of interest are $(\mathbb{N},+)$ or $(\mathbb{Z}/k\mathbb{Z},+)$ or $(\mathbb{Z}/k\mathbb{Z},.)$.

QUESTION : Has anyone seen this before?

I like to call them mo-morphisms ( phonetically it repeats as in the right hand side of the equation).

The motivation comes from the function $f(n) = n \mod 10$ ( any base will do ) on $(\mathbb{N},+)$. Another example on $(\mathbb{N},.)$ is $f(n):= rad(n)$. The product of the primes dividing $n$ (each prime counted once). It is easy to see that $rad(x.y)=rad(rad(x).rad(y))$.

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    $\begingroup$ Any such function induces a commutative semigroup structure $\tilde *$ on $f({\mathbb S})$ by $f(x) \tilde * f(y) := f(x*y) = f(f(x) * f(y))$, and the restriction of $f$ to $f({\mathbb S})$ obeys the simpler law $x \tilde * y = f(x) \tilde * f(y)$; conversely, these axioms imply the original property. When ${\mathbb S}$ is a group, the latter condition is equivalent to $f(x) = x \tilde * a$ for some order two element $a$ of $(f({\mathbb S}), \tilde *)$. $\endgroup$
    – Terry Tao
    Commented Sep 23, 2022 at 16:59
  • $\begingroup$ @Joseph at start I had the involutive axiom in (N,+) a lot but not all functions obey the idempotency. $\endgroup$
    – Jérôme JEAN-CHARLES
    Commented Sep 23, 2022 at 19:48
  • $\begingroup$ @Terry Ok for associativity , for Z/kZ,: I have the solution for the additive but not yet for the multiplicative case your observation help clarifying. $\endgroup$
    – Jérôme JEAN-CHARLES
    Commented Sep 23, 2022 at 19:57
  • $\begingroup$ The operation $*$ does not satisfy the identity f(xyz)=f(f(x)f(y)f(z)) in the case when $\mathbb{S}=\mathbb{Z}_2$, $*$ is addition modulo $2$ (XOR), and f(x)=x+1 mod 2 (NOT). If we added the assumption that $f=f^2$, then we would satisfy the identity f(xyz)=f(f(x)f(y)f(z)) though. $\endgroup$
    – Joseph Van Name
    Commented Sep 23, 2022 at 20:53
  • $\begingroup$ It seems to me like the way to get these is take a congruence on your semigroup and a set of representatives and let f be the map taking an element to its representative. $\endgroup$
    – Benjamin Steinberg
    Commented Sep 23, 2022 at 21:00

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