Take a functor $G:\mathbb D\to \mathbb C $, and call $Y:\mathbb D\to [\operatorname{\mathbb D,Set]}:D\mapsto\operatorname{Hom}_{\mathbb D}(-,D)$ the Yoneda functor. Consider the bifunctor $\operatorname{Hom}_{\mathbb D}(-,G-):\mathbb C \times \mathbb D\to \operatorname {Set}$, contravariant in the first variable and covariant in the second one. Intuitively, this bifunctor carries the same data of the functor $YG$; if one takes another $G':\mathbb D\to \mathbb C $ such that $\operatorname{Hom}_{\mathbb D}(-,G'-)$ is naturally isomorphic (in both variables) to $\operatorname{Hom}_{\mathbb D} (-,G-)$, one basically is saying that $YG\cong YG'$ (naturally); since $Y$ is fully faithful, $G\cong G'$ (naturally). Can anyone help me to prove rigorously what I said above (in particular where I wrote intuitively)? In class we just said that the it is a consequence of Yoneda.
P.S. If it can be useful, this argument came up in a proof that the right adjoint of a functor is uniquely determined modulo natural isomorphisms, observing that, given $F\dashv G,G':\mathbb D \to \mathbb C$, there are natural bijections $\operatorname{Hom}(-,G-)\cong \operatorname{Hom}(F-,-)\cong \operatorname{Hom}(-,G'-).$ I even read the proof of this fact in the book of Mac Lane (it's the corollary $\operatorname{IV.1.1}$), and I understand it, but I can't use it to solve my question.
This is a standard type of argument, but it is rarely spelled out. That makes it difficult for new people, it took me a while before I understood what and the rest follows by Yoneda means.
The crucial fact you need is that faithful functors reflect commutativity of diagrams. Of course all functors do preserve commutativity. The Yoneda embedding $Y: \mathcal D \to [\mathcal D^{op}, \mathbf {Set}]$ is fully faithful. So if for example $Ya \circ Yb = Yc \circ Yd\circ Ye$ holds in the category of functors, then it follows that $a\circ b = c\circ d\circ e$ is true in $\mathcal D$. This is because $Y$ is injective on homsets. Also fully faithful functors reflect isomorphisms.
Your functors $G$ and $G'$ should be of type $\mathcal C\to \mathcal D$. I assume you like to show that if \begin{align} \mathcal D(d,Gc) \cong \mathcal D(d,G'c) \end{align} naturally in both variables, then $G$ is naturally isomorphic to $G'$. This is the difficult direction. So assume you have a natural isomorphism as above. Then in particular \begin{align} \mathcal D(-,Gc) \cong \mathcal D'(-,G'c) \end{align} are naturally isomorphic for fixed $c$. By Yoneda's lemma any such isomorphism is of the form $\mathcal D(-,\alpha_c)$ for a unique morphism $\alpha_c: Gc \to Gc'$ in $\mathcal D$. The family of morphisms $\alpha_c$ we obtain this way assemble into the natural isomorphism $\alpha: G\Rightarrow G'$ we are trying to construct. We need to show that $\alpha_c$ are isomorphisms and that $\alpha$ is natural. Note that $Y\alpha_c = \mathcal D(-,\alpha_c)$ is an isomorphism by construction. The Yoneda embedding is fully faithful and therefor reflects isomorphisms. It follows that each $\alpha_c$ is an isomorphism. Next we check naturality. Take any morphism $f:x \to y'$ in $\mathcal C$. We need to show that \begin{align}\require{AMScd} \begin{CD} Gx @>{Gf}>> Gy\\ @V{\alpha_x}VV @VV{\alpha_y}V \\ G'x @>{G'f}>> G'y \end{CD}\end{align} But the Yoneda embedding reflects commutativity since it is faithful! So it is enough to apply $Y$ to the diagram and to check that the resulting diagram \begin{align}\require{AMScd} \begin{CD} \mathcal D(-,Gx) @>{\mathcal D(-,Gf)}>> \mathcal D(-,Gy)\\ @V{\cong}VV @VV{\cong}V \\ \mathcal D(-,G'x) @>{\mathcal D(-,G'f)}>> \mathcal D(-, G'y) \end{CD}\end{align} commutes in the functor category. But this is just the condition that the isomorphism we started with is natural in the second variable.
I hope it helps. Those arguments are all the same, you will get used to them.