Let $\mathcal{C}$ be a (sufficiently complete and cocomplete) closed monoidal category with internal hom $[-,-]$. Let $F : \mathcal{A} \to \mathcal{C}$ be a functor obtained as the left Kan extension of $F' : \mathcal{B} \to \mathcal{C}$ along a fully faithful functor $i : \mathcal{B} \to \mathcal{A}$. If $G : \mathcal{A} \to \mathcal{C}$ is another functor I can construct the composite $$ [F, G] : \mathcal{A}^\mathrm{op} \times \mathcal{A} \to \mathcal{C}^\mathrm{op} \times \mathcal{C} \to \mathcal{C}. $$ Since the value of $F$ on $\mathcal{A}$ depends only on the value of $F'$ on $\mathcal{B}$, it is reasonable to assume that the value of $[F, G]$ on $\mathcal{A}^\mathrm{op} \times \mathcal{A}$ depends only on the value of $[F', G]$ on $\mathcal{B}^\mathrm{op} \times \mathcal{A}$. In fact, thanks to the formula for left Kan extensions, $$ [F(a_1), G(a_2)] \simeq \left[ \operatorname*{colim}_{(b \to a_1) \in i/a_1} F'(b), G(a_2) \right] \simeq \operatorname*{lim}_{(b \to a_1) \in i/a_1} [F'(b), G(a_2)]. $$ This exhibits $[F, G] : \mathcal{A}^\mathrm{op} \times \mathcal{A} \to \mathcal{C}$ as a right Kan extension of $[F', G] : \mathcal{B}^\mathrm{op} \times \mathcal{A} \to \mathcal{C}$ along $\mathcal{B}^\mathrm{op} \times \mathcal{A} \to \mathcal{A}^\mathrm{op} \times \mathcal{A}$.

Is it possible to take this a step further and compute $[F, G]$ in terms of $[F', G \circ i]$? That is, how closely can I mimic the formula $\mathrm{Nat}(F, G) \simeq \mathrm{Nat}(F', G \circ i)$ (where $\mathrm{Nat}$ denotes natural transformations of functors) in my category $\mathcal{C}$ by using the internal hom instead of the hom of functors?