Warning: This is an incredibly stupid question. This comment seems to address the answer, but I want to make sure since I was not able to find this issue discussed anywhere else, and the comment does not seem to address this question directly.
The usual definition of adjunction starts with functors $F: D \to C$ and $G: C \to D$, then says that $$\operatorname{Hom}_C (F(\cdot), X) \cong \operatorname{Hom}_D(\cdot, G(X)) \,, $$ where the equivalence is natural in $X$, and that $$\operatorname{Hom}_C (F(Y), \cdot) \cong \operatorname{Hom}_D(Y, G(\cdot)) \,, $$ where the equivalence is natural in $Y$.
Question: Does this make sense strictly speaking? Namely I thought that the contravariant Hom functor $$\operatorname{Hom}_C(\cdot, X) $$ had "signature" $C^{op} \to Set$, but the target category of $F$ is $C$, not $C^{op}$.
Thus I don't see how $\operatorname{Hom}_C(F(\cdot), X)$ is well-defined, i.e. how can you compose a functor $D \to C$ with one $C^{op} \to Set$ and get a functor $D^{op} \to Set$? Shouldn't the first functor be $D^{op} \to C^{op}$? (E.g. like $F^{op}$.)
So is it (more) correct technically speaking to say that the adjunction conditions (replacing $F$ with $F^{op}$) are:
$$\operatorname{Hom}_C (F^{op}(\cdot), X) \cong \operatorname{Hom}_D(\cdot, G(X)) \,, \quad \text{and} \quad \operatorname{Hom}_C (F^{op}(Y), \cdot) \cong \operatorname{Hom}_D(Y, G(\cdot)) \,?$$