From Mac Lane:
I'm trying to prove the question at the bottom true, but I'm not exactly understanding how to jump between the different levels of abstraction.
If $B$ and $C$ are categories with one object each (say $B$ and $C$), then to satisfy the definition of a natural transformation it must be the case that if $g : B \rightarrow B$ is an arrow in $B$ (or an element of $B$) then $T(g) \circ \tau_B = \tau_B \circ S(g)$, where $\tau_B : S(B) \rightarrow T(B)$ is the natural transformation for each object (only one) in $B$. It seems obvious that the above equality should imply the conjugacy relationship, but I'm not exactly sure where to bridge the gap between the "category with one element called a group" and the "group" itself.
Is the correct way of thinking this?: Because the "group" $B$ with homomorphisms $T,S$ satisfies the axioms for the "category with one element called a group" and because we have the naturality relationship between the functors of the two "categories with one element called a group", we can replace the equation $T(g) \circ \tau_B = \tau_B \circ S(g)$ (that is in terms of arrows and objects) with their respective elements and products?
There aren't really different levels of abstraction here. Morphisms from the one object of the category $C$ to itself literally are elements of the group $C$. Composition of such elements literally is the group operation. So what you call $\tau_B$ literally is some element of the group $C$, as are $T(g)$ and $S(g)$, and your equation literally just says $T(g)\tau_B=\tau_B S(g)$ for these elements. There's not even any "replacing" to do; the morphisms in the category literally are elements of the group.