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Let us denote the covariant functor $C\rightarrow Mor(A,C)$ from $\mathcal{C}\rightarrow Sets$ by $h^A$ (We are assuming $\mathcal{C}$ is a locally small category). For a given $A,B \in \mathcal{C}$ We know there is a bijection between natural transformations $h^A \rightarrow h^B$ and $Mor(B,A)$ by mapping $\eta$ to $\eta_A(id_A)$. In fact, even more is true. Natural isomorphisms between $h^A$ and $h^B$ correspond to the isormorphisms in $Mor(B,A)$. This allows us to identify the subcategory of natural isomorphisms between $h^A$ and $h^B$ as a subset of $Mor(B,A)$.

Now let's take it one step further. Given a covariant functor $F:\mathcal{C}\rightarrow Sets$, we have a bijection between natural transformations $h^A\rightarrow F$ and $F(A)$ (by Yoneda's lemma). Is there a way to identify the set of natural isomorphisms between $h^A\rightarrow F$ as some subset of $F(A)$ (as we did above)?

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A satisfactory characterization is unlikely. The reason that you get a nice characterization (i.e., natural iso correspond to isos) in the case of $h^A\to h^B$ is that both domain and codomain are of the same type, and thus Yoneda works both ways. But, when you turn to general $h^A\to F$, for which Yoneda works, you are now interested in knowing when will there be an inverse natural transformation $F\to h^A$, and here Yoneda does not work.

Also, since $F(A)$ is nothing but a set, it is impossible to characterize those elements in it that correspond under Yoneda to natural isos without looking outside of the particular value $F(A)$. So, you must really look at the case at hand and can't expect to get a characterization for free, by so called general abstract nonsense. It's where you would stop using general abstract nonsense and start analyzing what's going on in the particular case you are looking at.

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Thanks for your comment. It really explains why a nice characterization for my first example works and why this doesn't work for general situations. –  Sergio Da Silva Mar 6 at 19:12

Yes, this is the set of universal elements of $F$ (on $A$). It is non-empty iff $F$ is representable by $A$. Let me spell this out: $x \in F(A)$ is universal iff for every other $y \in F(B)$ there is a unique $f : A \to B$ such that $y=F(f)(x)$. (Of course this is just a tautology, but finding universal elements is something which really depends on the specific choice of $F$ and may be very hard, so that in this generality nothing more can be said about this.)

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isn't the definition of universal element simply saying that the corresponding natural transformation is a natural iso? –  Ittay Weiss Mar 2 at 1:18
    
This is the name for elements of $F(A)$ corresponding to representations of $F$, but is it well understood how to find universal elements? –  Sergio Da Silva Mar 2 at 1:19
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@Sergio: This question is too broad. Finding universal elements is really doing mathematics. At Ittay: Yes. –  Martin Brandenburg Mar 2 at 1:20
    
@MartinBrandenburg :I can assume then that finding universal elements is not an easy task and is the subject of research? Thanks for your comments. –  Sergio Da Silva Mar 2 at 1:24
    
Yes. Essentially the question is how to determine whether a functor is representable. In rather algebraic situations, we may use Freyd's criterion. In more geometric situations, it is much more subtle. One of the most famous examples is probably Grothendieck's proof that the Hilbert scheme is representable. –  Martin Brandenburg Mar 2 at 1:25

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