I've been struggling for quite a while to understand why a vector space is considered to be "canonically embedded" into its double dual, but not its dual. As has been remarked in many other places, the distinction between whether an (iso-)morphism is "natural" can often seem vague and unintuitive. For me particularly, I think that part of the problem is that this sort of statement seems to run entirely counter to something I was taught early in Abstract Algebra as a Profound and Fundamental Lesson: "Isomorphic structures are exactly the same in all respects. When two things are isomorphic, all the things that can be said about one carry over verbatim to the other. There is no distinction between them." However, moving into more abstract linear algebra, a sort of about-face is being made, and now we are making the distinction of effectively saying, "My isomorphism is better than yours." In order to justify this apparent contradiction, the argument is typically made that the (iso-)morphism into the double dual does not require any "choices", while any embedding into the dual will require some "choice" to be made. However, this seems... unconvincing. So what if you can jury-rig a bilinear form out of whatever embedding/isomorphism I pick? Do we really have to pay attention to that? Again, this seems rather vague and unintuitive.
To make the argument more precise then, it is claimed that the ultimate answer lies in that Fountain of Eternal Truth - Category Theory. More specifically, it is claimed that the fact that there is a natural transformation from the identity functor on vector spaces to the double-dual functor justifies the claim that the embedding into the double-dual is "natural", while the fact that there is no such transformation between the identity and the dualizing functor shows that any such embedding into the dual is "not natural". This is elucidated beautifully in this thread. However, I claim that this is still not the final nail in the coffin of doubt. More specifically, I do not understand how natural transformations actually express the idea that a construction is (quotation marks) "natural".
How does this business with commuting diagrams make precise the idea that the embedding of a vector space into its double dual is "natural"? How does the implication that any association with the dual is "not natural" stem from a theorem saying that a certain collection of diagrams will never commute?
Another thing to note, that took me a little by surprise, is that the content of these arguments depends not only on the construction of the dual and double dual spaces, but also on this other construction called the transpose, which associates a linear map $f^*: F^* \to E^*$ to every linear map $f: E \to F$. So the fact that a map between a space and its dual is not "natural" also depends on the fact that we define an association between linear maps, that we package together with the dual operation to form the dualizing functor; however, this association between linear maps seems rather external to the association between a vector space and its dual. This is also bothering me.
I will not deny that the transpose operation does seem like a very natural thing to pair along with the dual operation, but what does seem odd is that the intrusiveness of this transpose operation should make or break the "naturalness" of something strictly between a vector space and its dual - honestly, who ordered that? Why can't I concoct some other association between linear maps - one that's covariant, at that, and package that together with the dual space to make something that admits a natural transformation from the identity? Note however that this would also break the "theorem" that a vector space is canonically embedded in its double dual, so this sort of train of thought is a double-edged sword.
Ultimately, I feel that I don't understand natural transformations in general very well; this example is really just the biggest one that sticks out to me and the one that I care about the most. I may post another question about the general case of understanding natural transformations, depending on how well this one goes and also whether I can manage to formulate it in a manner that seems intriguing and not simply lost and confused. At any rate, I look forward to any potential answers and would greatly appreciate whatever illumination you may be able to provide.
It seems to me that there are two possible meanings (close to each other).
One is that such isomorphism $**$ is defined via very simple and "expected" means. Another word commonly used for this is canonical. The definition $(v,w):=w(v)$ for $w\in V^*$ identifies $v$ with an element of $V^{**}$ and this does not depend on any additional structure on $V$, such as a metric. In this sense, it is "natural": you pair elements of $V^*$ with elements of $V$, so you can consider it the other way around as a pairing between elements of $V$ and $V^*$.
Another meaning is, as you say, the categorical. This basically says that not only can you apply $**$ to spaces but also to linear maps and the corresponding diagram commutes. That is, you can identify $f: X\to Y$ with $f^{**}: X^{**}\to Y^{**}$ (again, the identification goes via simple and expected means). As before, the functorial definition of $**$ does not depend on any additional structures such as metrics or scalar products.
These two meanings often go hand-in-hand: if something has a simple and expected definition (or a complicated one, but satisfying simple axioms), usually it can be converted into something categorical. It seems to me that if one wants to highlight the categorical meaning, (s)he uses the word natural, if one wants to highlight the simple and expected thing, (s)he often uses the word canonical.
But I'm not sure if this answers your question because I guess that you are aware of all of this.
Extension + edit: a small attempt to give some intuition of why natural transformation are "natural". Consider a finite-dimensional vector space $V$ and two its bases $\mathcal{B}_1$ and $\mathcal{B}_2$. Let $\mathcal{V}$ be a category with only one object $V$ and morphisms $Mor(V,V)$ being all linear transformations; let $\mathcal{R}$ be a category with one object $\Bbb R^n$ and morphisms being all linear transformations. You can define two functors $F$ and $G$ from $\mathcal{V}$ to $\mathcal{R}$ that express a linear transformation as a matrix wrt. the coordinates $\mathcal{B}_1$ resp. $\mathcal{B}_2$. A natural transformation between $F$ and $G$ assigns to the object $V$ in $Obj(\mathcal{V})$ the morphism $x\mapsto C^{-1}x$ of $\Bbb R^n$ (an element of $Mor(\Bbb R^n, \Bbb R^n)$), where $C$ is the transition matrix from base $\mathcal{B}_1$ to $\mathcal{B}_2$. This morphism is just the coordinate transformation in $\Bbb R^n$.
The fact that it is a natural transformation just reflects that any linear map $f: V\to V$ (an element of $Mor(V,V)$) gives rise to the commutative diagram \begin{array}{ccc} \Bbb R^n & \stackrel{F(f)}{\to} & \Bbb R^n \\ \downarrow_{C^{-1}} && \downarrow_{C^{-1}} \\ \Bbb R^n & \stackrel{G(f)}{\to} & \Bbb R^n \\ \end{array} or equivalently, \begin{array}{ccc} \Bbb R^n & \stackrel{M}{\to} & \Bbb R^n \\ \downarrow_{C^{-1}} && \downarrow_{C^{-1}} \\ \Bbb R^n & \stackrel{C^{-1}MC}{\to} & \Bbb R^n \\ \end{array} where $M$ is the matrix expression of $F(f)$. In physics, this corresponds to a change of observer: observer $\mathcal{B}_2$ will just "see" a vector $C^{-1}x$ and/or "use" the matrix $C^{-1}MC$ whenever observer $\mathcal{B}_1$ "sees" the vector $x$ and "uses" the matrix $M$. But they both see the same "real object". In this sense, the natural transformation is "natural".