I am reading and learning some basic category theory. I found the following problem on Categories for the Working Mathematician (I understand that it may not be the best learning source for a beginner, but the way Lane defines everything is so fascinating):
If $C, B$ are two groups regarded as categories with one object each and $S, T$ are functors, show that there exists a natural transformation if and only if $S$ and $T$ are conjugate; i.e., there exists $h\in B$ with $Tg=h(Sg)h^{-1}$ for all $g\in C$.
Here is the "if" part. Since $Tg=h(Sg)h^{-1}$, we have $T(g)\circ h=h\circ S(g)$. Therefore $h$ defines a natural transformation. However, I am not quite sure what the components of this natural transformation are. Since $h$ is an element in $B$, we can make $h$ be a component at $B$. But how can we define the component at $C$ by $h$?
I am even more confused when I get to the only if part. Suppose the following diagram commutes,
We have $T(\tau_C\circ g)=\tau_B\circ S(g)$ for all $g$ in $C$ (I know this is incorrect because the aimed "shape" is $T(g)\circ T(\tau_C)=\tau_B\circ S(g)$, so please correct my seriously false understanding). How can we go from this equation to the conjugacy?
Thanks in advance. I couldn't find a similar question on MathSE but please let me know if this post is a duplicate.
First note that the natural transformation $\tau:S\to T$ is indexed by the objects of the category $C$. Let $X$ denote the only object of $C$. Then for every morphism $g:X\to X$ in $C$ we have a commutative square$\require{AMScd}$ \begin{CD} S(X)@>\tau_X>>T(X)\\ @VS(g)VV@VVT(g)V\\ S(X)@>>\tau_X>T(X) \end{CD} where $\tau_X$ is the only component of $\tau$. If $Y$ is the only object of $D$, then $Y=S(X)=T(X)$ and putting $h=\tau_X$ the previous diagram become: \begin{CD} Y@>h>>Y\\ @VS(g)VV@VVT(g)V\\ Y@>>h>Y \end{CD} that's $S(g)h=hT(g)$.
In other words, the natural transformation $\tau$ has one component only (because the category $C$ as one object only) and this component is the morphism $h$ which satisfy $S(g)h=hT(g)$.