How is $\mathsf{Set}^{C^\mathsf{op}} \cong (\mathsf{Set}^C)^\mathsf{op}$? This is subtly referenced in Riehl's Category Theory in context; namely, a corollary of the Yoneda lemma for locally small categories is that the evaluation bifunctor is naturally isomorphic to $\text{Hom}(y-, -)$.
In showing this, the isomorphism in the title is implicitly used, and I've had no luck in proving this. Are there any pointers/tips for showing this is the case? Thanks in advance.
 A: This is not true.
To see this, consider for $C$ the one-object, one-morphism category, so that $D^C ≅ D$ for every category $D$.
On the one hand we have
$$
  ( \mathsf{Set}^C )^{\mathrm{op}} ≅ \mathsf{Set}^{\mathrm{op}} \,.
$$
On the other hand we have $C^{\mathrm{op}} ≅ C$ and thus
$$
  \mathsf{Set}^{(C^{\mathrm{op}})}
  ≅
  \mathsf{Set}^C
  ≅
  \mathsf{Set} \,.
$$
But the categories $\mathsf{Set}$ and $\mathsf{Set}^{\mathrm{op}}$ are not equivalent.
(In $\mathsf{Set}$ there exists a morphism from the initial object to the terminal object, but not the other way around.
In $\mathsf{Set}^{\mathrm{op}}$ the situation is reversed.)
What is true, however, is that
$$
  (D^C)^{\mathrm{op}} ≅ (D^{\mathrm{op}})^{(C^{\mathrm{op}})}
  \tag{$\ast$}
$$
for any two categories $C$ and $D$.
This is due to the following two observations:

*

*A functor from $C$ to $D$ is the same as a functor from $C^{\mathrm{op}}$ to $D^{\mathrm{op}}$.
Hence $(D^C)^{\mathrm{op}}$ and $(D^{\mathrm{op}})^{(C^{\mathrm{op}})}$ have the same objects.


*Let $F, G \colon C \to D$ be two functors and let $α \colon F \Rightarrow G$ a natural transformation.
This means that the components $α_X \colon F(X) \to G(X)$ are morphisms in $D$ such that the square
$$
  \require{AMScd}
  \begin{CD}
    F(X)         @> α_X >>   G(X)       \\
    @V F(f) VV               @VV G(f) V \\
    F(Y)         @> α_Y >>   G(Y)
  \end{CD}
$$
commutes for every morphism $f \colon X \to Y$ in $C$.
This is a commutative diagram in $D$.
If we regard $F$ and $G$ as functors $F', G' \colon C^{\mathrm{op}} \to D^{\mathrm{op}}$, then we can express the above square equivalently as the diagram
$$
  \require{AMScd}
  \begin{CD}
    F'(X)        @< α_X <<   G'(X)       \\
    @A F'(f) AA              @AA G'(f) A \\
    F'(Y)        @< α_Y <<   G'(Y)
  \end{CD}
$$
in $D^{\mathrm{op}}$.
Therefore, $α$ becomes a natural transformation from $G'$ to $F'$.
This leads to the occurrence of $(-)^{\mathrm{op}}$ in the left-hand side of $(\ast)$.
