Naturality of exponential objects in cartesian closed categories A category with finite products $C$ is cartesian closed if for each $X \in C$ the functor $- \times X \colon C \to C$ has a right adjoint $(-)^X$. This is to say that for each $Y,Z \in C$ we have bijections
$$
\hom(Y \times X, Z) \simeq \hom(Y,Z^X) \tag{1}
$$
that are natural both in $Y$ and $Z$.
Now, a map $f \colon X \to X'$ induces a natural transformation $1 \times f \colon - \times X \Rightarrow - \times X'$ and so we have a natural map
$$
\hom(-,(-)^{X'}) \simeq \hom(-\times X', -) \xrightarrow{(-\times f)^\ast} \hom(- \times X,-) \simeq \hom(-,(-)^{X}).
$$

Does this define a functorial assignment $X \mapsto (-)^X$?

Fixing $Y$, the above yields a natural transformation $\hom(-,Y^{X'}) \to \hom(-,Y^{X})$ which by Yoneda comes from a map $\eta_Y \colon Y^{X'} \to Y^{X}$. This looks natural on $Y$, but so far I haven't succeeded in proving so.
And in case the assignment is indeed functorial:

Is the adjunction-induced ismorphism on the hom-sets also natural in $X$? In other words, is $(1)$ a natural isomorphism of functors $C^{op} \times C^{op} \times C \to \mathsf{Set}$?

 A: $\require{AMScd}$
Let's say you have a morphism $g : Y\to Z$ in addition to $f : X \to X'$. By Yoneda, the square
$$
\begin{CD}
Y^{X'} @>f^*>> Y^X \\ 
@Vg_*VV @VVg_*V \\
Z^{X'} @>>f^*> Z^X
\end{CD}
$$
(where $f^*$ is precomposition, $g_*$ is postcomposition) commutes if and only if the square of natural transformations
$$
\begin{CD}
\hom(-,Y^{X'}) @>>> \hom(-,Y^X) \\ 
@VVV @VVV \\
\hom(-,Z^{X'}) @>>> \hom(-,Z^X)
\end{CD}
$$ from which you obtained the former commutes. This second square is isomorphic to the square of natural transformations
$$
\begin{CD}
\hom(X'\times-,Y) @>>> \hom(X\times -,Y) \\ 
@VVV @VVV \\
\hom(X'\times -,Z) @>>> \hom(X\times -,Z)
\end{CD}
$$
This is commutative if and only if for every object $A$ the square
$$
\begin{CD}
\hom(X'\times A,Y) @>>> \hom(X\times A,Y) \\ 
@VVV @VVV \\
\hom(X'\times A,Z) @>>> \hom(X\times A,Z)
\end{CD}
$$  commutes. It does -even naturally in $A$-.
RE the last question: yes, there is a isomorphism of functors
$$
\hom(-_1\times-_2,-_3)\cong \hom(-_1, -_3^{-_2}) \cong \hom(-_2, -_3^{-_1})
$$ (with obvious meaning of the placeholders). See the notion of adjunction of two variables for a more general look on this.
A: *

*For the first question : is $X\mapsto (-)^X$ a functor?

Yes, the assignation that you ahve defined makes this map into a functor. The only thing to check is that it preserves the identity and the composition. I will denote $\alpha_X : \operatorname{hom}(-,(-)^X)\to\hom(-\times X,-)$ the natural iso, to prove this. First we can prove that the assocation you have defined preserves the identity. Pluging in $f=\operatorname{id}$ yeilds
$$ \operatorname{hom}(-,(-)^X)\simeq \operatorname{hom}(-\times X,-) \overset{(-\times\operatorname{id})^*}\to \operatorname{hom}(-\times X,-)\simeq \operatorname{hom}(-,(-)^X)$$
This simplifies to $\alpha_x^{-1}\circ\alpha_X$, which is indeed the identity. Let's show that it also preserves the composition, considering $f : X \to X'$ and $g : X' \to X''$, You ahve the following equalities
\begin{align*}
\alpha_X^{-1}\circ(-\times g\circ f)^* \circ \alpha_{X''} &= \alpha_{X}^{-1}\circ(-\times f)^*(-\times g)^*\circ \alpha_{X''} \\
&=(\alpha_x^{-1}\circ(-\times f)^*\circ\alpha_{X'})\circ(\alpha_{X'}^{-1}\circ(-\times g)^*\circ\alpha_{X''})
\end{align*}
If you look at these expressions straight, they are exactly the fact that your defined action on functors preserves compositions

*

*For your second question : is $\eta_Y : Y^{X'}\to Y^X$ natural?

Well I believe this is exaclty the Yoneda lemma? I don't know what phrasing of this lemma you use, so I will use the following (where $\operatorname{Nat}$ is the set of natural transformations) : There is a natural iso
$$ \operatorname{Nat}(\operatorname{hom}(-,A),\operatorname{hom}(-,B)) \simeq \operatorname{hom}(A,B)$$
So for me the naturality is included in the statement of the lemma. You could verify by hand very easily though, so let's do the first part of that in your case: Considering a natural transformation $\alpha : \operatorname{hom}(-,Y^{X'}) \to \operatorname{hom}(-,Y^X)$, the map $\eta_Y$ defined this way is equal to $\eta_Y=\alpha(\operatorname{id}_{Y^{X'}})$. With this expression, you should be able to check the naturality, by using the naturality of $\alpha$.

*

*Regarding your last question : Do we have a functor $C^{op}\times C^{op} \times C \to Set$?

I am not sure which functor you want to consider, but if you consider the functor $\operatorname{hom}(-\times -,-)$, then yes. You can see that this is a functor because it is the composition of two functors $-\times -$ and $\operatorname{hom}(-,-)$. Similarly, the answer is also yes for $\operatorname{hom}(-,(-)^X)$
