The number of polynomials on a finite group, II

Question

This question is follow up of this MO-post.

First let us recall the necessary definitions.

A function $f:X\to X$ on a group $X$ is called a polynomial if there exists $n\in\mathbb N$ and elements $a_0,\dots,a_n\in X$ such that $f(x)=a_0xa_1x\cdots xa_n$ for all $x\in X$.

Let $\mathrm{Poly(X)}$ be the set of all polynomials on a group $X$. It is a submonoid of the monoid $X^X$ of all self-maps of $X$.

Observe that each polynomial $f$ on a commutative group $X$ is of the form $f(x)=ax^n$ for some $a\in X$ and $n\in \mathbb N$, which implies that $|\mathrm{Poly}(X)|=|X|\cdot\exp(X)$ for each finite commutative group $X$.

Here $\exp(X):=\min\{n\ge 1:\forall x\in X\;\;(x^n=1)\}$ is the exponent of the group $X$.

Since any group $X$ acts effectively on $\mathrm{Poly}(X)$ by left (or right) shifts, the cardinal $|\mathrm{Poly}(X)|$ is divisible by $|X|$.

Problem 1. Is $|\mathrm{Poly}(X)|$ divisible by $|X|\cdot\exp(X)$ for every finite group $X$?

The cardinality of the monoid $\mathrm{Poly}(X)$ was calculated by Peter Taylor for all non-commutative groups $X$ of cardinality $|X|<24$.

Problem 2. Is $|\mathrm{Poly}(X)|$ divisible by $|X|^2$ for every non-commutative finite group $X$?

Peter Taylow observed that the answer to Problem 2 is negative for the group $D_{12}=GAP(12,4)$ with $|\mathrm{Poly}(D_{12})|=648=2^3\times 3^4$. So, only Problems 1 and 3 remain open.

The calculations of Peter Taylor show that the following problem has affirmative answer for all finite groups of cardinality $<24$:

Problem 3. Let $X$ be a finite group and $p$ be a prime number dividing $|\mathrm{Poly}(X)|$. Is $p$ a divisor of $|X|$?

Remark 1. The answer to Problems 2 and 3 are affirmative for finite simple groups since $\mathrm{Poly}(X)=X^X$ for any non-commutative simple finite group $X$, see the answer of @YCor to this MO-question.

Answer 1

This is an answer to problem 1 and problem 3:

As noted in the comments, $\operatorname{Poly}(G)$ with pointwise multiplication is a group for finite $G$. Consider the homomorphism $\operatorname{Poly}(G)\to G$ given by evaluating at $1$. It is surjective (as seen by constant polynomials), and the kernel contains an element of order $\operatorname{exp}(G)$, the identity polynomial. So the kernel has order divisible by $\operatorname{exp}(G)$, and thus $\operatorname{Poly}(G)$ has order divisible by $\operatorname{exp}(G)\cdot |G|$.

Let me also answer problem 3! If $|\operatorname{Poly}(G)|$ is divisible by $p$, then there is an element $f\in \operatorname{Poly}(G)$ of order $p$. This means all values of $f$ have order dividing $p$, and at least one is nontrivial. So $G$ must have an element of order $p$, thus $p$ divides $|G|$.