I'm aware of the fact that there exists a natural isomorphism between a vector space $V$ and its double dual $V''$; what I'm asking is the following: let's define an application $F:V\to V'$ as $F(v)=v^T$ (the image of a vector is the application of the dual which corresponds to multiplying to the transpose of said vector); now for every $v\in V$ we can find $F(v)\in V'$ and vice versa, therefore $F$ is an isomorphism. Now I believe that could be considered a natural isomorphism since whatever are the basis for $V$ and $V'$, the coordinates of some vector with respect to said basis are unique and therefore to one vector in $V$ with respect to some basis $\mathscr{B}_V$ corresponds one and only application in $V'$ with respect to some basis $\mathscr{B}_{V'}$, even if $\mathscr{B}_{V'}$ is not the dual basis of $\mathscr{B}_V$.
Why isn't this a natural isomorphism? It seems to me I don't need to chose a basis for either $V$ or its dual in order to make this work...
If you're stil not convinced by the non-naturality of transposition, you'll surely agree on something more geometric, that there's no natural inner product.
Or even simpler, there's no natural length (norm) defined on all vectors in all spaces - doubling any length (norm) makes it a just as valid norm.
But any identification between $U$ and its dual $U^*$, $\eta:U\to U^*$ defines a scalar product on $U$ $$\langle u,v\rangle = \eta(u)(v).$$
If there was a natural isomorphism between $U$ and $U^*$, that would automatically define a scalar product and hence a norm (length) by $\|u\|=\sqrt{\langle u,u\rangle}$, i.e. turn all vector spaces into inner product spaces, which they initially are not :)