I am doing exercises from the Tom Leinster's Basic Category Theory and I want to check the answer to the following exercise:
Let $G$ be a group. For each $g\in G$, there is a unique homomorphism $\phi:\mathbb{Z}\to G$ satisfying $\phi(1)=g$. Thus, elements of $G$ are essentially the same thing as homomorphisms $\mathbb{Z}\to G$. When groups are regarded as one-object categories, homomorphisms $\mathbb{Z}\to G$ are in turn the same as functors $\mathbb{Z}\to G$. Natural isomorphism defines an equivalence relation on the set of functors $\mathbb Z\to G$, and, therefore, an equivalence relation on $G$ itself. What is equivalence relation, in purely group-theoretic terms?
My answer is that this relation can be caracterized in the following way: two elements $g,g'$ are in relation if there is automorphism $\alpha$ of $G$ such that $\alpha(g)=g'$.
Am I correct and is there a special name for this equivalence relation?
The automorphism $\phi$ should be inner so that $F = \phi \circ G$.
To see this, note that a natural transformation $\eta$ is a family of maps indexed by objects in $\mathbb Z$, which has only one object, so it is a single moprhism in $G$ (the target category, but as groups, a single element) so that for all $a \in \mathbb Z$ (or in other words, every morphism), we have that $G \circ \eta=\eta \circ F$, or $F=\eta^{-1} G \eta$, or $F=\phi \circ G$ for some inner automorphism.