How to make precise the meaning of naturality for a specific natural isomorphism? What I know: (1) If $V$ is a finite dimensional vector space then there is a natural isomorphism from $V$
to its double dual $V^{**}$. (2) There is no natural isomorphism from $V$ to its
dual $V^*$. (3) If we consider finite vector spaces with an inner product, then
there is a natural isomorphism from $V$ to $V^*$.
Intuitively, naturality can in all these 3 scenarios be understood as without need for some additional structure.
Naturality can also be made more precise in all 3 scenarios in a categorical sense:

*

*Define suitable categories (eg: vector spaces, vector spaces with inner products)

*Define functors (dualization, double dualization, identity functor)

*Check for a natural transformation between these functors.

The intuitive concept of naturality in linear algebra can be nicely translated into categorical language.
According to Wikipedia, Symmetric Tensor there exists a natural isomorphism between the vector space of symmetric tensors of order $d$ over an $n$-dimensional vector space and the dual of the vector space ${\mathbb K}[X_1, \ldots, X_n]$ of homogeneous polynomials of degree $d$ in $n$ variables over $\mathbb K$.
What I do not know: How would I formulate naturality in this situation in
a categorical sense?
My specific problem: I am not aware between which functors I should look for a natural transformation. I am looking for a full, precise definition of functors making
clear all the details of types, variances and more.
 A: I think there is a categorical description which captures some of the issues you are alluding to. It is based on my reading of Riehl,$\textit{Category Theory in Context }$ Chp2. Here is the setup:
Fix a field $\mathbb{K}$ and let $Vect_{\mathbb{K}}$ be the category of vector spaces over $\mathbb{K}$, with linear morphisms  and $Comm_{\mathbb{K}}$ commutative, associative, unital algebras over $\mathbb{K}$ with k-algebra morphisms. For $V\in Vect_{\mathbb{K}}$:
Let $\mathcal{T}(V) = \bigoplus_{k=0}^{\infty} V^{\otimes k}$.
Let $\mathcal{S}(V)$ be the symmetric algebra,
($\mathcal{S}(V) = \mathcal{T}(V)/\mathcal{I}$, where $\mathcal{I}=\langle x\otimes y - y \otimes x\rangle$).
Let $U: Comm_{\mathbb{K}} \rightarrow Vect_{\mathbb{K}}$ be the functor which forgets the algebra structure.
Let $G_n: Comm_{\mathbb{K}} \rightarrow Sets$ be the functor defined by
$G_n(A) = \{(a_1,\dots,a_n) : a_i \in A\} $, with obvious action on morphisms.
Finally, define functor $F_V :Comm_{\mathbb{K}} \rightarrow Sets$ by $F_V(A) = Hom_{Vect}(V, U(A))$.
Both $F_V$ and $G_n$ are covariant in $A$, are representable functors and have universal objects,namely, $\mathcal{S}(V)$ for $F_V$ and $\mathbb{K}[X_1,\dots,X_n]$ for $G_n$. I.e. we have natural bijections:
\begin{align*}
& Hom_{Comm}(\mathcal{S}(V),A)  \simeq Hom_{Vect}(V,U(A)) = F_V(A) & (1) \\
& Hom_{Comm}(\mathbb{K}[X_1,\dots,X_n],A)  \simeq G_n(A) & (2) 
 \end{align*}
The bijection in (1) is given by: $f\in  Hom_{Vect}(V,U(A)), f:V\rightarrow A \leftrightarrow \tilde{f}:(v_1, \dots ,v_n)\rightarrow f(v_1)\dots f(v_n) \in A$. $\tilde{f}:V\times \dots \times V \rightarrow A$ is multilinear and induces an algebra morphism $\mathcal{S}(V)\rightarrow A$. For (2), the bijection is given by $(a_1,\dots ,a_n) \leftrightarrow f:\mathbb{K}[X_1,\dots ,X_n]\rightarrow A$, where $f(X_i)=a_i$.
For (1), the bijection is natural in both $V$ and $A$, i.e. $\mathcal{S}$ is left adjoint to $U$.
Here is the point:
Let $n=\text{Dim}(V)$. Then there is a bijection between $\textit{sets}$ $F_V(A) \leftrightarrow G_n(A)$, as follows: fix an ordered basis $(v_1,\dots,v_n)$ for $V$. Then every $(a_1,\dots,a_n)\in G_n(A)$ defines a vector space homomorphism $V\rightarrow A$ by $v_i \rightarrow a_i$ and vica versa. So by the Yoneda lemma, this isomorphism in the category of $Sets$ implies an isomorphism between $\mathcal{S}(V)$ and $\mathbb{K}[X_1,\dots,X_n]$ in the $Comm_{\mathbb{K}}$ category. Furthermore, for fixed $V$, considering $F_V$ and $G_n$ as functors $Comm_{\mathbb{K}} \rightarrow Sets$, the bijection is natural. A different choice of basis will conjuagte the isomorphism of $\mathcal{S}(V)$ and $\mathbb{K}[X_1,\dots,X_n]$ by an automorphism of $V$.
Like many categorical arguments (at least for me) this either elucidates or obfuscates the issues. I hope you find it helpful.
