Using Yoneda lemma to get isomorphisms Let’s assume we have two functors $F:C\longrightarrow D$ and $G:C\longrightarrow D$ such that the functor $Hom_D(F(-),-)$ is naturally isomorphic to $Hom_D(G(-),-)$. Can we get from this that F is naturally isomorphic to G? For example, let’s assume we have a functor $F:C\longrightarrow D$ such that we have a natural bijection $Hom_D(F(X\times Y), Z)\simeq Hom_D(F(X)\times F(Y),Z)$. Then, if the first claim be true, can we use that to show F preserves binary products? 
 A: The answer is definitely yes to the first question. The second is a bit more subtle.
The first question
One proof goes like this:
For each $x\in C$, the natural bijection $D(Fx,-)\simeq D(Gx,-)$ gives an isomorphism 
 $\phi_x : Fx\xrightarrow{\sim} Gx$ (corresponding to $1_{Gx}$ on the right) by the Yoneda lemma. 
Naturality of $\phi$ comes from the fact that the original bijection was natural in $x$.
Suppose $f:x\to y$ is a morphism in $C$, then 
naturality of the original bijection implies that $(Ff)^*\phi_y = \phi_y\circ Ff$ corresponds to $(Gf)^*1_{Gy} = Gf = (Gf_*)1_{Gx}$,
and we know that $(Gf)_*1_{Gx}$ corresponds to $(Gf)_*\phi_x=Gf \circ \phi_x$.
Therefore $Gf\circ\phi_x$ and $\phi_y\circ Ff$ both correspond to $Gf$, so they must be equal, which implies that $\phi$ is natural.
The second question
What we mean when we say that $F$ preserves products is that if 
$(X\times Y, (\pi_X,\pi_Y))$ is a product in $C$, then 
$(F(X\times Y), (F\pi_X,F\pi_Y))$ is a product in $D$.
Note that there is a canonical morphism 
$$(F\pi_X,F\pi_Y) : F(X\times Y)\to F(X)\times F(Y),$$
meaning the morphism induced by the universal property of the product whose 
components are $F\pi_X$ and $F\pi_Y$. 
To say $F$ preserves products is equivalent to saying that this canonical morphism is always an isomorphism for all products.
It's not obvious that having a natural isomorphism $D(F(X\times Y),Z)\cong D(F(X)\times F(Y),Z)$ natural in $X$, $Y$, and $Z$ together (assuming choices of functorial products) forces the canonical morphism to be an isomorphism.
I'm kind of leaning towards believing that its false, but I don't have a counterexample, and perhaps naturality adds more than I think it does.
