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 $Aut^2(C_8) \simeq S_3$).

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 $n$.

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 $F^n(C_1) \ltimes N_n$.

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 $Aut^2(C_8) \simeq S_3$).

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 $Aut^2(C_8) \simeq S_3$).