An adjunction $F\dashv G$ gives rise to an adjunction $G^\ast \dashv F^\ast$ I'm working on this exercise (source):

The hint refers to this theorem:

But I don't see how exactly to use it. I'm interpreting the hint in the following way. Since $F\dashv G$, there are natural transformations $\eta, \epsilon$ as in the theorem, and they satisfy the triangle identities. Our job is to modify those triangular diagrams somehow (possibly, by applying some functors) to get the triangular identities for the adjunction $G^\ast \dashv F^\ast$. But first of all, I don't see how to modify the original triangular diagrams, and secondly, we still need to prove that the functors $N:1_{[\mathscr A,\mathscr C]}\to F^\ast\circ G^\ast$ and $E: G^\ast\circ F^\ast\to 1_{[\mathscr B,\mathscr C]}$ (the functors that should satisfy the triangular identities) exist, which needs to be proven separately.
 A: By Theorem 2.2.5, the adjuction $F \dashv G$ is equivalent to a unit $\eta: 1_\mathscr A \implies GF$ and a counit $\epsilon: FG \implies 1_\mathscr B$ satisfying the triangle identities
(1):  and
(2): .
Define $\eta^* : 1_{\mathscr S^\mathscr A} \implies F^*G^*$ with $\eta^*_X = X\eta$ for any functor $X \in \mathscr S^\mathscr A$. Define $\epsilon^*:G^*F^* \implies 1_{\mathscr S^\mathscr B}$ with $\epsilon^*_Y= Y\epsilon$ for any functor $Y \in \mathscr S^\mathscr B$.
By the naturality of $\eta$ and the functoriality of any functor $X \in \mathscr S^\mathscr A$, $\eta^*$ is a natural transformation. Similarly, $\epsilon^*$ is also a natural transformation. Therefore, it remains to show that $\eta^*$ and $\epsilon^*$ satisfy the triangle identities.
Let $X$ be a functor in $\mathscr S^\mathscr A$. Whiskering (2) with $X$, we obtain the following commutative diagram due to the functoriality of $X$
 
Since $X \circ GFG = G^*(XGF) = G^*F^*(XG) = G^*F^*G^*X$, $X\eta G = G^*(X\eta) = G^*\eta^*_X$ and $(XG)\epsilon = \epsilon^*_{G^*X}$, the above commutative diagram can be rewritten as
(3): .
Now let $Y$ be a functor in $\mathscr S^\mathscr B$. Whiskering (1) with $Y$, we obtain the following commutative diagram due to the functoriality of $Y$

Since $Y \circ FGF = F^*(YFG) = F^*G^*(YF) = F^*G^*F^*Y$, $YF\eta = \eta^*_{F^*Y}$ and $Y\epsilon F = F^*(Y\epsilon) = F^*\epsilon^*_Y$, the above commutative diagram can be rewritten as
(4): .
Assembling (3) for all $X \in \mathscr S^\mathscr A$ and (4) for all $Y \in \mathscr S^\mathscr B$ shows that $\eta^*$ and $\epsilon^*$ satisfy the triangle identities. By Theorem 2.2.5, this gives an adjunction $G^* \dashv F^*$.
