So let $F: \mathcal {A} \mapsto \mathcal{B}, G: \mathcal {B} \mapsto \mathcal{A}$ be two functors. Clearly, those can be understood as morphisms of the $\mathbb{CAT}$ - category of all categories.
My textbook says that $G \circ F = 1_{\mathcal{A}}, F \circ G = 1_{\mathcal{B}}$ is unreasonably strict, much less useful definition. Even though my current knowledge is not sufficient to understand "unreasonable strictness", I trust it. Right after textbook proudly says that following definition is far better: $$G \circ F \cong 1_{\mathcal{A}}, F \circ G \cong 1_{\mathcal{B}}$$
And that's the point where I get completely stuck. Obviously, $G \circ F$ is a map in the $\mathbb{CAT}$; therefore, it is a functor (I guess there is something else here - it is always endofunctor: it maps category to itself; not sure whether this fact is important). But natural transformation is the only way one able to define and measure "isomorphism between functors" (with requirements that both do have same domain and codomain, which is met here). However, natural transformation does not live in the $\mathbb{CAT}$, it lives in an appropriate "functor category", where functors are objects rather than morphisms. I don't quite understand such a "mental jump" from the $\mathbb{CAT}$ to a something completely different.
So in order to figure it out, I try to draw a simplest possible schema. $$(G \circ F): \mathcal{A} \mapsto \mathcal{A}$$ $$1_{\mathcal{A}}: \mathcal{A} \mapsto \mathcal{A}$$ $$\lambda: (G \circ F) \mapsto 1_{\mathcal{A}}$$ $$\forall X \in \mathcal{A}: \lambda_X \text{ is an isomorphism ... WITH WHAT?! }$$ Last statement blows my head off! $\lambda_X$ is a morphism from a $((G \circ F)X)$ to a $((1_{\mathcal{A}})X)$. Does it require those two to be isomorphic in a regular sense?
Kindly asking someone smart to explain what am I missing here avoiding sophisticated examples from other branches of mathematics?
Thanks in advance.