Natural transformations in a group category I've been thinking of a group as a category with one object such that all the arrows are invertible. I have a few questions:
First, am I correct in thinking that a natural transformation between group homomorphisms $f,g: A \rightarrow B$ would be elements $b\in B$ such that $bf = gb$?
This next question may be stupid. I know that homomorphisms induce continuous maps of the classifying spaces, and natural transformations induce homotopies between those maps. Since the classifying space $\mathbb{Z}$ has a the homotopy type of a circle, do homomorphisms $f:\mathbb{Z} \rightarrow G$ give us maps that are essentially paths? And can we define a product between such homomorphisms that agrees with the product in the fundamental group?
 A: I don't know much about classifying spaces, but as for the last question, $\mathbb{Z}$ is the free group on one generator; consequently, $\text{Hom}(\mathbb{Z}, G)$ is canonically isomorphic to $G$. This is more than a set, though; it has a group structure which is natural in $G$, so by the Yoneda lemma it follows that $\mathbb{Z}$ has a cogroup structure inducing the group structure on $G$. The cogroup operation $\mathbb{Z} \to \mathbb{Z} \ast \mathbb{Z}$ (where $\ast$ denotes the free product) sends a generator of $\mathbb{Z}$ to the product of the two generators of the two copies of $\mathbb{Z}$, and you can verify that it induces the usual group structure on $G$. If you get stuck on this, see this blog post.
(Note that the pointwise product of homomorphisms is not a homomorphism in general unless $G$ is abelian.)
(Note also that $S^1$ is a cogroup in the pointed homotopy category, and the fundamental group functor sends this cogroup structure to that of $\mathbb{Z}$.)
A: 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).
A: (Re: 1) Yes, definitely (so, for example, endofunctor equivalent to identity = inner automorphism).
(Re: 2) Sure. $\operatorname{Hom}(\mathbb Z;G)=G$ (isomorphism is just $f\mapsto f(1)$). And isomorphism $\pi_1(BG)=G$ is a group isomorphism.
