Bifunctoriality stronger than functoriality in each variable?

Is it correct to say that bifunctoriality is a stronger condition than "functoriality in each variable"?

More precisely, suppose that $$\mathcal A, B, C$$ are categories and I have $$F \colon \mathcal A \times \mathcal B \to \mathcal C$$. Bifunctoriality of $$F$$ means that $$F$$ is a functor from the product category $$\mathcal A \times \mathcal B$$ to $$\mathcal C$$. On the other hand, functoriality in each variable means that $$F(-, B) \colon \mathcal A \to \mathcal C$$ is a functor for each fixed $$B \in \mathcal B$$, and likewise $$F(A, -) \colon \mathcal B \to \mathcal C$$ is a functor for each fixed $$A \in \mathcal A$$. Given $$A \to A'$$ and $$B \to B'$$, bifunctoriality requires the following diagram to commute, but I don't see why functoriality in each variable should, too: $$\require{AMScd}$$ $$\begin{CD} F(A,B) @>>> F(A',B)\\ @VVV & @VVV\\ F(A,B') @>>> F(A',B') \end{CD}$$ My question has the following motivation: to define adjoint functors $$F$$ and $$G$$, some authors (e.g. Vakil's algebraic geometry, Altman/Kleiman's commutative algebra) will state that there should be a "natural bijection" $$\operatorname{Hom}_{\mathcal B}(F(A),B) \to \operatorname{Hom}_{\mathcal A}(A, G(B))$$ for every $$A \in \mathcal A$$ and $$B \in \mathcal B$$. They then go on to explain that by "natural" they mean some commutative squares that essentially amount to naturality in each variable. I was wondering why they don't just come out and say that $$\operatorname{Hom}_{\mathcal B}(F(-), -)$$ and $$\operatorname{Hom}_{\mathcal A}(-, G(-))$$ should be naturally isomorphic bifunctors on $$\mathcal A^{\text{op}} \times \mathcal B$$ (which is the approach taken on the Wikipedia page for adjoints), and I thought perhaps this latter requirement is actually a stronger one that happens not to matter for $$\operatorname{Hom}$$.

Edit: We can find easy examples by thinking about one-object categories. If $$G, H$$ are two groups and $$BG, BH$$ the corresponding one-object categories, then a bifunctor $$BG \times BH \to C$$ is an action of $$G \times H$$ on an object of $$C$$, but a map which is a functor in each variable separately is an action of the free product $$G \ast H$$ on an object of $$C$$.
However, this is irrelevant to your motivation because $$\text{Hom}(F(a), b)$$ and $$\text{Hom}(a, G(b))$$ are already bifunctors. Since every morphism in the product category $$A^{op} \times B$$ is canonically a composite of a morphism in $$A$$ and a morphism in $$B$$, to check that a collection of maps $$\eta_{a, b} : \text{Hom}(F(a), b) \to \text{Hom}(a, G(b))$$ assemble into a natural isomorphism of bifunctors you only have to check it with respect to morphisms in $$A$$ and then morphisms in $$B$$.