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Let $mathbb{F}$ be a field. The category of matrices $mathbf{Mat}$ has $mathbb{N}_0$ as class of objects and $mathrm{hom}(n,m)=mathbb{F}^{ntimes m}$ for $n,minmathbb{N}_0$, with $mathrm{id}_n=mathbf{1}_n$ and $Bcirc A=Acdot B$ with the usual matrix multiplication whenever defined. The category of finite-dimensional vector spaces $mathbf{FinVect}$ has $mathbb{F}$-vector spaces as objects and linear maps as morphisms, where composition is the usual composition of functions and the identity morphism is the identity map. The categories $mathbf{Mat}$ and $mathbf{FinVect}$ are equivalent.

There is a contravariant duality functor $(-)^astcolonmathbf{FinVect}rightarrowmathbf{FinVect}$ with $V^ast=mathrm{Hom}(V,mathbb{F})$ for any vector space $V$ and $f^astcolon W^astrightarrow V^ast,,gmapsto gcirc f$ for any linear map $fcolon Vrightarrow W$. There also is a contravariant transpose functor $(-)^tcolonmathbf{Mat}rightarrowmathbf{Mat}$ where $n^t=n$ for $ninmathbb{N}_0$ and $A^t$ is the usual transpose of a matrix.

These two contravariant functors on equivalent categories are related by the fact that for finite-dimensional vector spaces $V,W$ with bases $mathscr{B},mathscr{C}$ respectively, we have that
$$(T^{mathscr{B}}_{mathscr{C}}(f))^t=T^{mathscr{C}^ast}_{mathscr{B}^ast}(f^ast),$$
where $T$ denotes the transformation matrix for the respective bases.

Is there a way to describe this relationship of functors in category-theoretical terms?

There's an obvious functor $text{Mat} to text{FinVect}$ which is an equivalence, but no obvious choice of inverse to this functor; choosing an inverse amounts to choosing, for each finite-dimensional vector space (note that there is more than a set's worth of such things!), a basis of that vector space. Not a fun operation to perform; we need something like the axiom of global choice to do it.

What can be done without making such choices is the following: there is an intermediate category which I'll just call $text{Base}$, the category of finite-dimensional vector spaces together with a choice of basis (morphisms are still just linear transformations). There is an obvious functor $F : text{Base} to text{FinVect}$ given by forgetting the basis, and an obvious functor $M : text{Base} to text{Mat}$ given by writing linear transformations with respect to a basis. Both of these functors are equivalences, but again the functor $F : text{Base} to text{FinVect}$ can't really be inverted obviously. See also anafunctor.

$text{Base}$ has a dualization endofunctor $D : text{Base} to text{Base}$ which takes duals of linear transformations and sends a pair $(V, B)$ of a vector space and a basis to the pair $(V^{ast}, B^{ast})$ of the dual vector space and the dual basis. Now what we can say is that the functors $F$ and $M$ intertwine this dualization functor and the other dualization functors, in the sense that we have not just natural isomorphisms but equalities

$$F circ D = (-)^{ast} circ F : text{Base} to text{FinVect}$$
$$M circ D = (-)^T circ M : text{Base} to text{Mat}.$$