Let $\textbf{A, X}$ be categories and $F:\textbf{X} \to \textbf{A}$ and $G: \textbf{A} \to \textbf{X}$. Then there is a map that takes an object in $\text{Arr}(\textbf{X}\times\textbf{A})$ (the arrow category) to a commutative diagram in $\textbf{Set}$ as in the following illustration. And vice versa.
where $S := \textbf{Set}$ so that $S(X,Y) \equiv \text{Hom}_{S}(X,Y)$. I did that in order to make a less visually-complex diagram. And where $\theta$ is the natural isomorphism $\text{Hom}(-,G-) \xrightarrow{\sim} \text{Hom}(F-, -)$.
That commutative $2\times1$ square (forall such $\xi, \alpha$) is in fact the very definition of natural isomorphism of homsets as in the standard definition of adjoint pair of functors, so I argue that $F,G$ are adjoint if and only if this map is defined for all objects in the category $\text{Arr}(\textbf{X}\times\textbf{A})$ i.e. the diagram shown commutes.
I'm wondering though,
Can we draw arrows back in 3D to another diagram in the image of above-mentioned map, that we'll call $H:\text{Arr}(X\times A) \to \text{The category of }2\times 1\text{ squares in } \textbf{Set}$, so that adjoint pairs $F,G$ will always induce, not only a map into these commutative rectangles, but also a functor into this category?
where morphisms are the obvious morphisms in $\textbf{Set}$ which form a 3D extrusion of the above-linked diagram.
What's neat about this construction is you could probably apply it to other categorical constructions as well.
It might be beneficial first to make the situation a bit more general. Suppose we have two functors $P_0,P_1 : \newcommand{\cA}{\mathbf{C}}\newcommand{\cB}{\mathbf{D}}\cA⇉\cB$ and a natural transformation $α : P_0→P_1$. In your situation, these two functors are $\hom(-,G-), \hom(F-,-) : \mathbf{X}^{\mathrm{op}} × \mathbf{A} → \mathbf{Set}$. For each morphism $f : x→y$ in $\cA$, there is the naturality square associated, the one you drew in the Quiver website (and in your situation it can be drawn as a $2×1$ square). The question is to put all of these squares into a coherent whole.
One way to do that is to view the natural transformation as a functor itself $\cA × I → \cB$ where $I$ is the "arrow category" with two objects $0$ and $1$ and one arrow from $0$ to $1$. The functor $\cA × I → \cB$ sends $(x,i)$ to $P_i(x)$. It sends an arrow $f : (x,i) → (y,i)$ to $P_i(f) : P_i(y) → P_i(x)$ and it sends $f : (x,0) → (y,1)$ to the composite of $P_0(f) : P_0(x) → P_0(y)$ and $α_y : P_0(y) → P_1(y)$, which is equal to the composite of $α_x : P_0(x) → P_1(x)$ and $P_1(f) : P_1(x) → P_1(y)$.
Now, another way but actually equivalent is to view our functor $\cA × I → \cB$ as a functor $\cA → [I,\cB]$ where $[I,\cB]$ is the category of functors $I→\cB$ (the arrow category of $\cB$, in other words). Instead of sending $(x,i)$ to $P_i(x)$, we send $x$ to $i↦P_i(x)$. I think the picture you have in your mind is $[I,-]$ applied to this functor, which then gives a functor $[I,\cA] → [I,[I,\cB]] = [I×I,\cB]$ (from arrows in $\cA$ to commutative squares in $\cB$).
In the end, we rather have a functor from $\operatorname{Arr}(\mathbf{X}^{\mathrm{op}} × \mathbf{A})$ (instead of $\operatorname{Arr}(\mathbf{X} × \mathbf{A})$) to the category of squares in $\mathbf{Set}$. If you want these $2×1$ squares instead of just squares, then instead of applying the $[I,-]$ construction to $\cA → [I,\cB]$, you can apply $[[2],-]$ where $[2]$ is the category with three objects $0$, $1$ and $2$, a morphism $0→1$, a morphism $1→2$ and a morphism $0→2$. We get a functor $[[2],\mathbf{X}^{\mathrm{op}} × \mathbf{A}] → [[2]×I,\mathbf{Set}]$ and we can precompose it with the functor $[I,\mathbf{X}^{\mathrm{op}} × \mathbf{A}] → [[2],\mathbf{X}^{\mathrm{op}} × \mathbf{A}]$ that decomposes an arrow as the canonical composite of the two coordinate arrows.