Why does the result you mention in your fourth paragraph not answer your question?

@EricWofsey How does that paragraph necessarily mean that out of all possible things that could be necessary to fix this, we need additional structure involving arbitrary choices?

Well, you haven't given any definition of what "we need additional structure involving arbitrary choices" means. It's not at all obvious how this should be defined rigorously. One perfectly reasonable definition would be the non-existence of a dinatural isomorphism. It is fine if you prefer a different definition, but you need to explain what your definition is.

@EricWofsey In an effort to make it a bit more concrete, how does the non-existence of such a dinatural transformation concretely show us that we need to pick an equivalent of an inner product? It's not me jumping to vaguely defined conclusions out of the blue; this is commonplace in literature and I'm just trying to get a proper understanding of what's going on.

I don't think it's true that you need to pick an equivalent of an inner product. You just need to pick something that is not determined by just the vector space structure (since if you had a definition that used only the vector space structure, it would be dinatural, at least with respect to isomorphisms).

Maybe your actual question is about my last parenthetical: why is "dinatural with respect to isomorphisms" a reasonable way to make "using only the vector space structure" precise?

So very strictly speaking, the mainstream statement "There is no canonical isomorphism between $V \to V^*$ because this requires the choice of (some equivalent of) an inner product" is incorrect? And the best we can say is that $V$, just by itself, won't support such an isomorphism to it's dual? Logically, this is quite different!

Well, it's close to correct. Note that an isomorphism $V\to V^*$ is easily seen to be equivalent to a nondegenerate bilinear form on $V$. So, you don't necessarily have the symmetry or positivity that is usually assumed of an inner product, but it's a similar concept.

@EricWofsey if all that we can infer is that $V$ by its structure alone can't support an isomorphism $V \to V^*$, how does looking at this mapping more generally as a nondegenerate bilinear form help?

Help with what?

With making the statement in question 'almost correct'

It's almost correct in that it would be correct if you replaced "inner product" with "nondegenerate bilinear form".

I think your focus on inner products is kind of misleading, though.

Even if defining an isomorphism V \to V^* required an inner product, that wouldn't show (on its own) that there is no canonical isomorphism.

After all, maybe there's a canonical inner product!

So, all it does is push the question from "is there a canonical isomorphism V\to V^*?" to "is there a canonical inner product on V?".

Maybe you find the latter question easier to answer for some reason, but there's no obvious reason it should be.

Similarly, the equivalence between isomorphisms V \to V^* and nondegenerate bilinear forms on V just says the question "is there a canonical isomorphism V\to V^*?" is the same as the question "is there a canonical nondegenerate bilinear form on V?".

It doesn't answer either question.

Is there a way to formalize the notion that there is no canonical inner product on V, apart from the observation that arbitrary choices of bases etc yield different equally candid inner products?

Sure: it's just the statement that there is no inner product on V that is invariant under all automorphisms of V.

Since an automorphism is just a change of basis, this is essentiially equivalent to the statement that different bases give different inner products.

So then this fact almost immediately also proves that there is no canonical isomorphism from V to V^* ?

Well, no, because an isomorphism from V to V^* is not the same thing as an inner product.

But if you replace "inner product" with "nondegenerate bilinear form", then yes.

There is also no nondegenerate bilinear form invariant under all automorphisms of V, is there?

Not in general.

There is for a few special vector spaces.

(And correspondingly, those vector spaces do have canonical isomorphisms to their duals.)

This is discussed in the answer to the question you linked.

Is the fact that no such dinatural transformation exist then in one to one correspondence with the fact that in general there are no nondegenerate bilinear forms invariant under all automorphisms of V?

Yes, if by "dinatural" you mean "dinatural with respect to isomorphisms".

Dinaturality with respect to all morphisms is a stronger condition (and doesn't really correspond to "canonical" well).

Wow, this wasn't intuitive at all, thank you! Is there actually any quick and intuitive way to see that in this case dinaturality with respect to isomorphisms exactly corresponds to the absence of a nondegenerate bilinear form invariant under all automorphisms of V?

Yes. Just write down the definitions and chase what they mean.

It's really the exact same statement, just rearranged.

If you have an automorphism f of V, you can write down the square that must commute for your isomorphism V \to V^* to be dinatural with respect to f.

And that commuting square says exactly that the corresponding nondegenerate bilinear form is invariant under f.

This has been mind expanding. Again, thank you very much!

I would also more than happily accept an answer to the original question, involving going over the right way to approach the problem pure algebraically as in this chat; and how the algebraic approach directly corresponds to the category theoretical result with dinatural transformations.