For (1), it's worth doing the calculation explicitly, as it is the simplest possible situation in which to have a natural transformation, and thus useful for building intuition. A natural transformation $\eta:F\Rightarrow G$ between two functors $F,G:\mathcal C \to \mathcal D$ is a collection of morphisms $\eta_c\in \mathcal D(F(c),G(c))$ for each $c\in \mathcal C$ such that, then \eta_{c'}F(f)=G(f)\eta_c for every f:c\to c'
When $\mathcal{C}, \mathcal{D}$ have one object, $\eta$ is described by a single morphism in $\mathcal{D}$ which intertwines between $F$ and $G$. Switching notation back to the variables in the original equation, if $A$ and $B$ are groups and $f$ and $g$ are homomorphisms, then a natural transformation from $f$ to $g$ is a $b\in B$ such that $bf(a)=g(a)b$ for all $a\in A$.
You could shorten this to $bf=gb$, but to me, it isn't immediately clear what that would mean.
Because $B$ is a group, we can rephrase this slightly by writing $bf(a)b^{-1}=g(a)$, or $g=c_b \circ f$ where $c_b$ denotes the inner automorphism of $B$ coming from conjugation by $b$. In this case, every natural transformation is a homotopy equivalence, and the equivalence classes of are homomorphisms up to conjugation. I believe that there are connections between this and what you have asked in (2), but I do not currently have a complete answer to give.
If $f=g$, so that we can compose two natural transformations, we see that $\operatorname{End}(f)$ is the centralizer of $f(A)$ in $B$, that is, the elements of $b$ which commute with $f(a)$ for all $a\in A$. If we specialize further to the case that $A=B$ and $f=g=\operatorname{id}_A$, we have $\operatorname{End}(\operatorname{id}_A)=Z(A)$, the center of $A$. This generalizes slightly: if $A$ is an abelian category with one object (which is equivalent to being a ring), the natural transformations between the identity functor has a ring structure and will be the center of $A$ viewed as a ring. More generally, we can define the Bernstein center $Z(\mathcal C):=\operatorname{End}(\operatorname{id}_{\mathcal C})$, which for regular categories will be a monoid and which for enriched categories will have additional structure (I believe that it will be a monoid in the category you are enriching over).