Is the standard isomorphism $L(F^n,F^n)\to F^{n\times n}$ a natural transformation? Let $V$ be an $n$-dimensional vector space over a field $F$. I was wondering if the algebra isomorphism $$\Phi\colon L(F^n,F^n)\to F^{n\times n}$$ defined by
$$A(x)=\Phi(A)\cdot x\quad\forall x\in F^n$$
is in some sense a "natural transformation", e.g. as defined here.
 A: Here is an option, but first some notation:

*

*If $\textsf{C}$ and $\textsf{D}$ are categories, and $F : \textsf{C} \to \textsf{D}$ is a functor, then $F^{\rm op} : \textsf{C}^{\rm op} \to \textsf{D}^{\rm op}$ denotes the functor defined on objects and morphisms as same as $F$.

*If $F_1: \textsf{C}_1 \to \textsf{D}_1$ and $F_2 : \textsf{C}_2 \to \textsf{D}_2$ are arbitrary functors, then $$F_1 \times F_2 : \textsf{C}_1 \times \textsf{C}_2 \to \textsf{D}_1 \times \textsf{D}_2$$ denotes the functor defined by $(x_1,x_2) \mapsto (F_1(x_1),F_2(x_2))$ (for objects and morphisms).

*For a locally small category $\textsf{C}$ we have the two-sided represented functor $\operatorname{Hom}_\textsf{C} : \textsf{C}^{\rm op} \times \textsf{C} \to \textsf{Set}$ which sends a pair of objects $(x,y)$ in $\textsf{C}$ to the set $\operatorname{Hom}_\textsf{C}(x,y)$.

Now note that if $\textsf{C}$ and $\textsf{D}$ are locally small, every functor $F : \textsf{C} \to \textsf{D}$ induces a natural transformation $F^\flat : \operatorname{Hom}_\textsf{C} \to \operatorname{Hom}_\textsf{D} \circ \,(F^{\rm op} \times F)$ whose component at $(x,y)$ is the function $$\operatorname{Hom}_\textsf{C}(x,y) \to \operatorname{Hom}_\textsf{D}(F(x),F(y))$$ which sends $f : x \to y$ to $F(f) : F(x) \to F(y)$.
Finally, if $k$ is a field, in the special case in which

*

*$\textsf{C}$ is the category $\textsf{Mat}_k$ whose objects are the natural numbers $0,1,2,\dots$ and morphisms $n \to m$ are $m \times n$ matrices with entries in $k$,

*$\textsf{D}$ is the category $k\textsf{-Vect}$ whose objects are $k$-vector spaces and morphisms are linear maps, and

*$F$ is the functor $k^{(\_)} : \textsf{Mat}_k \to k\textsf{-Vect}$ which takes $n$ to $k^n$ and a matrix $A$ to the linear map $A\_$ (left-multiplication by $A$);

then we see that $F^\flat$ is a natural transformation whose component at $(n,m)$ is the function $$k^{m \times n} = \operatorname{Hom}_{\textsf{Mat}_k}(n,m) \to \operatorname{Hom}_{k\textsf{-Vect}}(k^n,k^m) = L(k^n,k^m)$$ given by $A \mapsto A\_$, so that the component at $(n,n)$ is the inverse of your map $\Phi$.
A: There are at least two problems in producing such natural transformation. First is basis for $F^n$. Second is how the domain and codomain of the linear map can be ensured to contain same field of scalars. But you can assume this and consider $A_{ij} = L(B_i)_j$, where $L : F^n \rightarrow F^m$ and $A : F^{n \times m}$. The intermediate vector $B_i$ is a inverse image of L for projection $A_i$ of a matrix $A$ to its i'th row vector, $B_i \in L^{-1}(A_i)$. Note using this projection in defining scalars of the matrix has serious self-reference problem, but assuming in abstract that the row vectors $B_i$ exist, will require existence of basis sufficient that it's possible to apply L to the row vector. It's possible to choose the formal expression for representation of vectors $B_i = \sum_{j \in [1,n]}{B_{ij}x_j}$ where $x_j$ are the basis elements and compute using the free vector space.
The natural transformation is a map taking $L$ to $A$, with $F$ (the field of scalars) kept abstract by both the functor and natural transformation. In detail
$α : L(-^n,-^m) \rightarrow (-^{n \times m}) : Fld \rightarrow Set$, defined by $α_F(L : L(F^n,F^m))_{ij}=L(B_i)_j$.
But as the above shows, the natural transformation would have to invent $B_i$ from thin air, only using $F$ and $L$.
