Understanding Equivalence of Categories An equivalence of two categories $\mathcal{C},\mathcal{D}$ consists of a pair of functors $F:\mathcal{C} \rightarrow \mathcal{D}$, $G:\mathcal{D} \rightarrow \mathcal{C}$ and natural isomorphisms $FG \rightarrow \mathrm{id}_{\mathcal{D}}$, $GF \rightarrow \mathrm{id}_{\mathcal{C}}$.
Now then, what is a natural isomorphism? Let $S,T:\mathcal{A} \rightarrow \mathcal{B}$ be equi(?)variant functors. A natural isomorphism $\tau:S \rightarrow T$ is a family of morphisms in $\mathcal{B}$, $\tau = (\tau_A:S(A)\rightarrow T(A))_{A\in \mathrm{ob}(\mathcal{A})}$, such that for any $f \in \mathrm{Hom}_{\mathcal{A}}(A,A')$ we have $\tau_{A'}S(f)=T(f)\tau_A$, and each $\tau_A$ is an isomorphism. (This is from Rotman's Introduction to Hom. Algebra.)
Finally, the actual question: What does this entail in the context of equivalence of categories (first paragraph)? My guess is the following: A natural isomorphism $\varepsilon:FG \rightarrow \mathrm{id}_\mathcal{D}$ means a family of morphisms in $\mathcal{D}$, $\varepsilon=(\varepsilon:FG(D) \rightarrow \mathrm{id}_\mathcal{D}(D)=D)_{D \in \mathrm{ob}(\mathcal{D})}$ such that for any $f \in \mathrm{Hom}_\mathcal{D}(D,D')$, $\varepsilon_{D'}FG(f)=f \varepsilon_D$ and each $\varepsilon_D$ is an isomorphism. And analogously for the other pair of functors.
My confusion lies in what functors are "compared" in the equivalence.
 A: It works pretty much like other isomorphisms in mathematics. For instance, if I want to show that two topological spaces are isomorphic, I want maps $f:X \rightarrow Y$ and $g:Y \rightarrow X$ such that $g \circ f = id_X$ and $f \circ g = id_Y$. Now, it would make sense to say that an equivalence of categories is a pair of functors $F: \mathcal{C} \rightarrow \mathcal{D}$ and $G: \mathcal{D} \rightarrow \mathcal{C}$ such that $G \circ F = id_\mathcal{C}$ and $G \circ F = id_\mathcal{D}$, where you compose functors as expected and $id$ is the identity functor.
But the problem is that in practice, wanting the two functors to be exactly equal is asking far, far too much, because in category theory, we usually don't care if two objects are exactly equal. If they're isomorphic, that's almost always good enough. So we relax the equality above to natural isomorphism. 
So for each object of $D$, we have that $FG(D)$ is isomorphic to $D$ via some isomorphism $\epsilon_D:FG(D) \rightarrow D$, and that all these isomorphisms are compatible with arrows in $D$, in the sense that if we go from $D$ to $D'$ via $f:D \rightarrow D'$, then $f \circ \epsilon_D = \epsilon_{D'} \circ FG(f)$. (You might even think of this as the map $FG(f)$ being "isomorphic" to the map $f$, in some sense.) So that part of the equivalance is like "global isomorphism" from all the objects and arrows spit out by $FG$ to all the original objects and arrows in $\mathcal{D}$. And similarly for the other composition.
