Is there a unique natural transformation of functors $t : F \to F'$ such that $t(U) = T(\text{Id}_F(U))$? Let $C, D$ be categories, $F$ and $F'$ covariant functors from $C$
to $D$.
For any natural transformation of bifunctors $T : \hom_D(−, F(−)) \to
\hom_D(−, F'(−))$ there is a unique natural transformation of functors $t :
F \to F'$ such that $t(U) = T(\text{Id}_F(U))$.
I tried using Yoneda to see the naturality of $t$, but it didn't work here. Any help would be appreciated.
 A: 
Let $J: A\to B$ be a fully faithful functor.
Then, for any category $E$, the functor $J_*: Fun(E,A)\to Fun(E,B), G\mapsto J\circ G$ is also fully faithful.

It's a very important lemma, which is not too hard to prove. Be careful though, in trying to prove it, that the same lemma does not hold if you replace "fully faithful" by "full", so to prove fullness you will need to use fully faithfulness of $J$.
Then consider $y: D\to Fun(D^{op},Set)$ the Yoneda embedding. By the Yoneda lemma, it is fully faithful, by the lemma, $y_* : Fun(C,D)\to Fun(C,Fun(D^{op},Set))$ is also fully faithful.
Howevern $Fun(C,Fun(D^{op},Set))\cong Fun(C\times D^{op},Set)$, so your functors $\hom_D(-,F(-)) $ and $\hom_D(-,F'(-))$ live in this category, more specifically they can be written as $y_*(F)$ and $y_*(F')$ respectively.
It follows that $T: y_*F\to y_*F'$ can be written as $y_*t$ for some $t: F\to F'$, which is automatically natural (and unique).
Now note that for a fixed $U\in C$, $T_{(-,U)}$ is a natural transformation $\hom_D(-,F(U))\to \hom_D(-,F'(U))$ induced in fact by $t_U$. The Yoneda lemma automatically implies that it must be satisfy $t_U= T_{(F(U),U)}(id_{F(U)})$.
In particular this definition of $t$ makes it natural (if that's what you were worried about), and of course unique (because well it gives a fixed definition of $t$) - and (to my mind, more importantly) an antecedent of $T$ through $y_*$.
