Let $F\dashv G:\sf C\to D$ be an adjunction (of locally small categories) and let $\sf I,J$ be small categories. Then $F,G$ induce by postcomposition an adjunction $F_*\dashv G_*:\sf C^J\to D^J$.
For this I should have a solution. We must prove that $\mathsf {C^J}(-,G_*-)\cong \mathsf {D^J}(F_*-,-)$; so take two functors $H$ in $\sf C^J$ and $K$ in $\sf D^J$. The data of a natural transformation $H\Rightarrow GK$ contain a morphism $Hj\to GKj$ for every $j\in\mathrm {ob}(J)$, and by $F\dashv G$ we get morphisms $FHj\to Kj$. The fact that, again by $F\dashv G$, one of these square commutes iff the other does proves that a set of morphisms $Hj\to GKj$ assembles in a natural transformation iff the corresponding $FHj\to Kj$ do, i.e. we have a bijection $\Psi_{H,K}:\mathsf{C^J}(H,GK)\cong \mathsf{D^J}(FH,K)$. $\require{AMScd}$ $$\begin{CD}
Hj @>>> GKj\\ @VHiVV @VGKiVV\\ Hj'@>>> GKj'
\end{CD}$$ $$\begin{CD}
FHj @>>> Kj\\ @VFHiVV @VKiVV\\ FHj'@>>> Kj'
\end{CD}$$ For the naturality in $K$ (in $H$ is analogous) take $\alpha:K\Rightarrow K'$ and $\phi:H\Rightarrow GK$. By construction, the components of $\Psi_{H,K'}(G\alpha\cdot \phi)$ are the transposes $\overline{G\alpha_j\cdot \phi_j}$, that are exactly $\alpha_j\cdot \bar \phi_j$, the components of $\alpha\cdot \Psi_{H,K}(\phi)$.
I have troubles in proving that $G^*\dashv F^*:\sf I^D\to I^C$ instead, where $F^*,G^*$ are induced by precomposition. Here the the exercise (4.4.iii on Riehl's CTC) suggests to work with the unit and counit. So take $H$ in $\sf I^C$ and $D$ in $\sf I^D$, plus a natural transformation $\alpha :H\Rightarrow KF$. For the unit $\eta^*_H:H\Rightarrow HGF$, we can choose $H\eta$, where $\eta$ is the unit of $F\dashv G$. Now we need a natural transformation $HG\Rightarrow K$, that we can choose as $K\epsilon\cdot \alpha G$. Thus, if these guesses are correct, we should prove that $(K\epsilon\cdot \alpha G\cdot H\eta)F=\alpha$, and that $K\epsilon\cdot \alpha G$ is unique in this sense. They wouldn't seem hard but I've been trying for a while without success, and I don't even see another possible choice for $\eta^*$. Any hint is appreciated, thanks.
Best Answer
You've already noticed that $\eta^*_H$ can be taken to be the natural transformation $i \mapsto H\eta_i: Hi\to G^* F^* H i = H G F i.$ For the counit $\epsilon^*$, take, unsurprisingly, $e^*_H$ to be $i \mapsto H\epsilon_i: F^* G^* Hi = H F G i \to H i.$
Now all you need to check the the triangle identities. But they follow directly from definitions, for example $\epsilon_{F^*H}^* \circ F^* \eta^*_H: F^*H \implies F^*G^* F^*H \implies F^*H$ is for every $i$ just $H\epsilon_{Fi} \circ H F \eta_i: HFi \to HFGFi \to HFi$, which is $1_{HFi}$ since this is $H$ applied to the triangle identity for $F \dashv G$ at the point $i$.