I am reading "Categories for the working mathematician" written by Mac Lane.
There is an exercise in natural transformation section: If $B$ and $C$ are groups (regarded as categories with one object each) and $S, T: B\to C$ are functors (homomorphisms of groups), show that there is a natural transformation $S\to T$ if and only if $S$ and $T$ are conjugate; i.e. iff there is an element $h\in C$ with $Tg=h(Sg)h^{-1}$ for all $g\in B.$
My attempt: Let $f$ be an endomorphisms from $Hom(B,B)$. If there is a natural transformation $\tau$, since there is only one component, the above diagram must be commutative. $\Rightarrow Tf\circ \tau=\tau\circ Sf$.(I) Let $f=1_B$, therefore we have $(1_C\circ \tau)(g)=(\tau\circ 1_C)(g)$.
(I): If in the question $g$ was a morphism $\in Hom(B,B)$, from equation (I) we have $Tg=\tau(Sg)\tau^{-1}$
Best Answer
Let's spell everything out. We view the group $B$ as a category with one object $*_B$, the only Hom-set is $\text{End}(*_B)=B$ and composition is given by the group law. Similar for $C$. If $S\colon B\to C$ is a functor, then $S$ consists of the following data: $S(*_B)=*_C$ and for each $b\in \text{End}(*_B)$, $S(b)\in \text{End}(*_C)$ such that $S(bb')=S(b)S(b')$.
A natural transformation $\eta\colon S\Rightarrow T$ consists of the following data:
The second condition is simply $$T(c)=\eta_{*_B}S(c)\eta_{*_B}^{-1}.$$ This is precisely saying that $S$ and $T$ are conjugate.