Classification of the functors on the category of cyclic groups

Question

Let $\mathsf{Grp}$ be the category of groups and let $\mathsf{Cyc}$ be the subcategory of cyclic groups.
As seen in the posts here and there (and their answers), a functor $F: \mathsf{Cyc} \to \mathsf{Cyc}$ is a very structured/restrictive notion, we are then lead to wonder whether there exists such a functor which is non-equivalent to the identity or the trivial functor, or if there is a such functor with $F(C_1) \not \simeq C_1$. As pointed out by Martin Brandenburg and Jeremy Rickard, $C_1$ is a retract of $F(C_1)$, so that $F(C_1)$ must be a retract of $F^2(C_1)$, and more generally, $F^n(C_1)$ is a retract of $F^{n+1}(C_1)$, which means that $F^{n+1}(C_1)$ is isomorphic to a semidirect product $F^n(C_1) \ltimes N_n$; now $F^{n+1}(C_1)$ is a cyclic group, so the semidirect product is in fact a direct product and moreover $gcd(|F^n(C_1)|,|N_n|) = 1$.

Question: What are the functors on the categroy of cyclic groups?

Remark: $Aut(-)$ is not such a functor because $Aut(C_8) \simeq C_2 \times C_2$ (and $Aut^2(C_8) \simeq S_3$).

In his answer, Neil Strickland provides examples of functors $F$ with $F(C_1) \not \simeq C_1$ and with $F^2(C_1) \not \simeq F(C_1)$, but with $F^3(C_1) \simeq F^2(C_1)$.

Bonus question: Is there a functor $F: \mathsf{Cyc} \to \mathsf{Cyc}$ such that $F^{n+1}(C_1) \not \simeq F^n(C_1)$ for all $n$?

Remark: If so, the sequence $(F^n(C_1))_n$ cannot be periodic (for $n$ large enough), because then (as shown above) $F^{n+1}(C_1) \simeq F^{n}(C_1) \times N_n$ with $|N_n|>1$ for all $n$.

Answer 1

I'll use additive notation, and I'll assume that you are only considering finite cyclic groups. Let $\mathbf{Cyc}_p$ be the category of cyclic $p$-groups. Given $i,j\geq 0$ we can define $Q(p;i,j)\colon\mathbf{Cyc}\to\mathbf{Cyc}_p$ by $Q(p;i,j)(A)=\{a\in p^iA\colon p^ja=0\}$. We can also define a constant functor $C(p;i)\colon\mathbf{Cyc}\to\mathbf{Cyc}_p$ by $C(p;i)(A)=\mathbb{Z}/p^i$. Now suppose we have a collection of functors $F_p$, one for each prime $p$, each of the form $Q(p;i,j)$ or $C(p;i)$, and that only finitely many of the functors $F_p$ are constant. Then the group $F(A)=\prod_pF_p(A)$ is cyclic for all $A$, so we get a functor $F\colon\mathbf{Cyc}\to\mathbf{Cyc}$. I don't know if that gives all possible functors, but it certainly gives a reasonably rich supply of them.

As a very specific example, the functor $F(A)=(A/2A)\times(\mathbb{Z}/3)$ is non-constant with $F(0)\neq 0$.

UPDATE: Here's a more exotic example. If $X$ is a based set of size $1$ or $3$, there is a unique group structure for which the basepoint is the identity. If $A$ is cyclic of order $1$ or $7$ then we can impose an equivalence relation with $a\sim a^2\sim a^4$ for all $a$, and then $A/\sim$ has size $1$ or $3$ with basepoint $0$ and so has a group structure. This construction gives a functor on the category of groups of order $1$ or $7$, and we can compose with $A\mapsto A/7$ to get a functor $\mathbf{Cyc}\to\mathbf{Cyc}_3$. I am sure that there are many variations on this theme.