What is the name of this identity relating the monoidal products of finite sets and finite-dimensional vector spaces? Note: This question is almost certainly a duplicate. Since I don't know the terminology involved, I couldn't find the original question. If someone can find the original question and link to it, then I don't mind if you mark this question to be closed.

Notation: Let $\mathfrak{S}$ denote the category of finite sets with arbitrary functions between them, let $\mathfrak{V}$ denote the category of finite dimensional real vector spaces, let $\mathfrak{C}$ denote the "category" of categories, let $\times$ denote the Cartesian product bifunctor $\mathfrak{S} \times \mathfrak{S} \to \mathfrak{S}$, let $\otimes$ denote the vector space tensor product bifunctor $\mathfrak{V} \times \mathfrak{V} \to \mathfrak{V}$, and let $\operatorname{Diag}$ denote the functor $\mathfrak{C} \to \mathfrak{C} \times \mathfrak{C}$ which sends a category $\mathcal{C}$ to the product category $\mathcal{C} \to \mathcal{C}$ and a functor $\mathfrak{F}$ to the bifunctor $\mathfrak{F} \times \mathfrak{F}$ whose "components" are just the original functor $\mathfrak{F}$.
Then as far as I can tell the following identity of functors is true:

$$\newcommand{\Hom}{\operatorname{Hom}} \Hom_{\mathfrak{S}}( \cdot, \mathbb{R}) \otimes \Hom_{\mathfrak{S}}(\cdot, \mathbb{R}) = \Hom_{\mathfrak{S}}( \cdot \times \cdot, \mathbb{R})$$

where the contravariant Hom functor $\Hom_{\mathfrak{S}}(\cdot, \mathbb{R})$ is interpreted as being $\mathfrak{S} \to \mathfrak{B}$.
In other words, letting $\mathfrak{F}$ denote $\Hom_{\mathfrak{S}}(\cdot, \mathbb{R})$, then as far as I can tell the following is true:

$$ \otimes \circ \operatorname{Diag}(\mathfrak{F}) = \mathfrak{F} \circ \times \,. $$
Question: What is the name of this identity?


Random thoughts: It's almost as if the functor $\mathfrak{F}$ is "commuting" with two different products.
And although I don't actually conceptually understand what a closed monoidal category is, I have read that $\mathfrak{S}$ is a closed monoidal category with tensor product $\times$ (Cartesian product), and that $\mathfrak{B}$ is a closed monoidal category with tensor product $\otimes$. So it seems like the functor $\mathfrak{F}$ is somehow "relating" the closed monoidal category structure of $\mathfrak{S}$ with the closed monoidal category structure of $\mathfrak{B}$, but if so I don't know what term to attach to it.
For example, $\mathfrak{F}$ obviously couldn't be a natural transformation between the tensor products $\times$ and $\otimes$ for example (which embarrassingly was the first idea that came to mind), since the two bifunctors $\times$ and $\otimes$ don't even have the same two categories as their "domain" and "codomain".
Finally, since $\mathfrak{F}$ is technically a $\Hom$ functor, and $\times$ and $\otimes$ are both tensor products of closed monoidal categories, my second thought was that somehow this must be related to the "tensor-Hom" adjunction that apparently exists within any closed monoidal category. However, that doesn't make sense either, since the "tensor-Hom" adhjunction (to the best of my knowledge) for a closed monoidal category only applies to the category's internal Hom functor.
But $\mathfrak{F} = \Hom_{\mathfrak{S}}(\cdot, \mathbb{R}) : \mathfrak{S} \to \mathfrak{V}$ isn't an internal Hom functor, neither for $\mathfrak{S}$ nor for $\mathfrak{B}$. I'm not sure it's exactly what is meant by an "external" Hom functor either (my understanding was that was just the term for the internal Hom functor of the category of sets, to distinguish it from the internal Hom functors of other categories), but $\mathfrak{F}$ is in some sense "external" to both $\mathfrak{S}$ and $\mathfrak{V}$. So even though it is a Hom functor, and even though it relates the tensor products of two different closed monoidal categories, this identity involving it seems unrelated to an "tensor-Hom" adjunctions.
Also, is this related to the Yoneda Lemma somehow? (If I'm being honest I don't actually conceptually understand that either, I just know that it has to do with the Hom sets of objects from some category which isn't the category of sets giving a "representation" of that category within the category of sets, or something like that, which allows one to use Hom sets reduce a lot of category theory to set theory, or something like that.)
 A: The word you’re looking for is strong monoidal functor, and the ingredient you’re missing is precisely the functor from sets to vector spaces, mapping a set $S$ to a vector space with basis $S$. Calling this free vector space functor $F$, what you have observed in effect is a natural isomorphism $F(S)\otimes F(T)\cong F(S\times T)$, which is the key property of a strong monoidal functor.
A: My thoughts.
First, I'm going to drop all the fraktur because I can't read it.
Let $\newcommand\Set{\mathbf{Set}}\newcommand\fin{{}_{\textrm{fin}}}\newcommand\Vect{\mathbf{Vect}}\Set$ be the category of sets, $\fin\Set$ the category of finite sets, and $\fin\Vect_k$ the category of finite dimensional vector spaces over a field.
You've defined a functor $F:\fin\Set\to\fin\Vect_k$ by $F(X)=\fin\Set(-,k)$, but unless $k$ is finite, this doesn't make sense. Rather you should define it as $\Set(-,k)$ instead (where you restrict the inputs to finite(!) sets). However, I think this perspective doesn't explain why it has the properties it has.
Instead, I think it's better to think of $F$ as the dual vector space of the free vector space functor.
First, an aside

(!) Let $F$ be the actual free vector space functor in this section.

Let's review what this means. Note that if we let $U:\Vect_k\to \Set$ be the forgetful functor, sending a vector space to its underlying set, then 
we have a natural isomorphism of sets
$$\Vect_k(F(X),V)\simeq \Set(X,U(V))$$
I.e., $F$ is left adjoint to the forgetful functor $U$.
Now what does this tell us? Well, let's recall two other adjunctions.
For $V,W\in \Vect_k$, note that $\Vect_k(V,W)$ is naturally also a $k$-vector space, 
the vector space of linear maps from $V$ to $W$.
Then we have adjunctions:
$$
\Vect_k(U\otimes V,W)\simeq \Vect_k(U,\Vect_k(V,W))
$$
and
$$
\Set(X\times Y,Z)\simeq \Set(X,\Set(Y,Z)).
$$
Then 
$$
\begin{align}
\Vect_k(F(X\times Y),W)
&\simeq \Set(X\times Y, UW)\\
&\simeq \Set(X,\Set(Y,UW)\\
&\simeq \Set(X,U\Vect_k(FY,W))\\
&\simeq \Vect_k(FX,\Vect_k(FY,W))\\
&\simeq \Vect_k(FX\otimes FY,W).
\end{align}
$$
Thus by Yoneda lemma, we have a natural isomorphism $F(X\times Y)\simeq FX\otimes FY$.
So what's going on?
Well, in this particularly nice example, we have two closed monoidal categories each enriched over themselves, and a natural functor $U:\Vect_k \to \Set$, so that we get a very nice adjunction, where 
$U\Vect_k(FX,V)\simeq \Set(X,UV)$.
However we can connect to more general results. $U$ is naturally a lax monoidal functor since the tensor product is defined by a bilinear map $UV\times UW\to U(V\otimes W)$.
And left adjoints of lax monoidal functors are apparently strong monoidal functors.
Returning to your functor
Let $D:\Vect_k\to \Vect_k $ be the dualization functor,
$DV=\Vect_k(V,k)$. Your functor is naturally the composite $DF$ restricted to finite sets,
since 
$$DF(X) = \Vect_k(FX,k) = \Set(X,Uk).$$
Now 
$$D(V\otimes W) 
\simeq \Vect_k(V\otimes W,k)
\simeq \Vect_k(V,\Vect_k(W,k)),
$$
and if $V$ is finite dimensional, this last is naturally isomorphic to 
$\Vect_k(V,k)\otimes \Vect_k(W,k)=DV\otimes DW$.
Then since $F$ preserves the tensor product, and $D(V\otimes W) \simeq DV\otimes DW$ for finite dimensional vector spaces, we get the result you've observed.
