Any $z \in \widehat{\mathbb{Z}} = \lim\_{n} \mathbb{Z}/n\mathbb{Z}$ defines an operation on all finite groups: if $G$ is a finite group and $g \in G$, say $g^n=1$, then map it to $g^{z\_n}$. This defines a map of sets $G \to G$, which is natural with respect to all homomorphisms of finite groups. The same works for torsion groups. And these are all operations, as can be seen from the Yoneda Lemma applied to the ind-representable forgetful functor $\mathbf{FinGrp} \to \mathbf{Set}$.


Of course, we get an actual homomorphism when $z = 0$ or $z = 1$. I believe that these are the only cases (right?).


**Question.** If $z \in \widehat{\mathbb{Z}} \setminus \{0,1\}$, is there always a finite group $G$ of some order $n$ such that $G \to G$, $g \mapsto g^{z\_n}$ is not a homomorphism? How can we prove this?


For $z \in \mathbb{Z} \setminus \{0,1\}$, we can argue as follows: In the free group $F = \langle a,b \rangle$ we have $(ab)^z \neq a^z b^z$. Since $F$ [residually finite](https://en.wikipedia.org/wiki/Residually_finite_group), there is a finite group $G$ and a homomorphism $\varphi : F \to G$ such that $\varphi((ab)^z) \neq \varphi(a^z b^z)$, and we are done. (Is there a more direct proof which *constructs* $G$ in this case?)


Equivalently, my question is how to prove $Z(\mathbf{FinGrp}) \cong \{0,1\}$, where $Z(-)$ is the [center](https://ncatlab.org/nlab/show/center) of a category. Equivalently, $Z(\mathbf{ProFinGrp}) \cong \{0,1\}$.



Classification of natural endomorphisms on finite groups
It suffices to look at symmetric groups, as these are generated by transpositions. Indeed, let $z \in \widehat{\mathbf Z}$ and suppose it induces a homomorphism $\phi\_z$ on $S\_n$ for all $n$ given by $g \mapsto g^{z\_n}$ if $g$ has order $n$. We will show that $z$ is either $0$ or $1$.


* If $z\_2 = 0 \in \mathbf Z/2\mathbf Z$, then each $\phi\_z \colon S\_n \to S\_n$ is a homomorphism killing all transpositions, therefore $\phi\_z$ is the trivial homomorphism. Thus $z\_n = 0 \in \mathbf Z/n\mathbf Z$ for all $n$, as $1 = \phi\_z(g) = g^{z\_n}$ for any element $g \in S\_n$ of order $n$.
* If $z\_2 = 1 \in \mathbf Z/2\mathbf Z$, then each $\phi\_z \colon S\_n \to S\_n$ is the identity on all transpositions, hence $\phi\_z$ is the identity. Thus $z\_n = 1 \in \mathbf Z/n\mathbf Z$ for all $n$, again since $g = \phi\_z(g) = g^{z\_n}$ for $g \in S\_n$ of order $n$. $\square$


