Natural isomorphisms between finite-dimensional vector spaces and their duals There are several questions on this already, and my question arose after reading the explanation given in the following:

In categorical terms, why is there no canonical isomorphism from a finite dimensional vector space to its dual?

We know a natural isomorphism $\eta:F \to G$ is defined as a natural transformation whose components are all isomorphisms in $\mathbf{D}$, where $F,G: \mathbf{C} \to \mathbf{D}$ are functors. What is the relation between isomorphisms in the category of vector spaces, which are just morphisms (and certainly not arrows between functors, it seems to me), and a "natural" isomorphism, i.e. the category-theoretic definition?
Also, in the answer to the quoted question, the author states that there is a natural isomorphism between a Euclidean vector space and its dual, and gives it as $V \mapsto V^*, v \mapsto \langle v,- \rangle$. What does this notation mean? And also, if there are natural isomorphisms between some finite-dimensional vector spaces, how can we "prove" that there are none in general, as the OP wants? It seems like the best we could do is a counterexample? Or am I missing a key distinction between Euclidean spaces and finite-dimensional spaces?
 A: A Euclidean space is not just a finite-dimensional vector space. It is a finite-dimensional vector space over the reals together with an inner product space. So it is an object of a very different category than mere vector spaces (over the reals). The extra structure gives in this case a particular choice for a mapping $V\to V^*$, namely using the inner product. In the absence of inner products, that is if all you have is a finite-dimensional vector space over the reals, then there is no canonical choice for a mapping $V\to V^*$ even though the two spaces have the same dimension. So, they are isomorphic but there is no canonical choice for any particular isomorphism.
Further, the construction $V\mapsto V^*$ is contravariant, so it does not even make sense to ask for a natural transformation between the identity functor on $\mathrm {Vect}$ and the (contravariant!) functor $-^*$. The double dual is covariant and so the question becomes meaningful and there is then a natural isomorphism between the functors.
You could account for the reversal of arrows and try to compare the identity functor and the contravariant dualisation. Once this is given precise meaning it can be shown that any assumed such transformation leads to a contradiction, so it does not exist.
A: 
What is the relation between isomorphisms in the category of vector spaces, which are just morphisms (and certainly not arrows between functors, it seems to me), and a "natural" isomorphism, i.e. the category-theoretic definition?

I assume that the most puzzling aspect of this is the fact that the term "isomorphism" is used for two different kinds of objects; one is a map between vector spaces, and the other is a family of such maps. In the most general sense, "isomorphism" can refer to any reversible structure preserving map. An invertible linear map between vector spaces is called an isomorphism because it preserves the "linear" structure of a vector space. In particular, the essential thing that makes an object a vector space is the ability to take linear combinations of elements, i.e. to ("nicely" and "meaningfully") refer to an object $ax + by$ for scalars $a,b$ and vectors $x,y$. $f:V \to W$ is an isomorphism, then this essential structure is preserved in the sense that
$$
f(ax + by) = af(x) + bf(y).
$$
In other words, taking a linear combination in the first space and then applying an isomorphism yields the same result as applying the isomorphism and then taking (the same) linear combination in the second space.
On the other hand, the essential thing that makes a map $F:\mathbf C \to \mathbf D$ a functor is the fact that $F(f \circ g) = F(f) \circ F(g)$ holds for all compatible morphisms $f,g \in \mathbf C$. A natural isomorphism $\eta:F \to G$ preserves this structure in the sense that for any morphism $f:X \to Y$ in $\mathbf C$, we have
$$
\eta_Y \circ F(f) = G(f) \circ \eta_X. 
$$
In other words, applying the first functor to a morphism and then applying the natural isomorphism yields the same result as applying the natural isomorphism and then applying the second functor to that same morphism.
A: $
\newcommand{\A}{\mathscr A}
\newcommand{\B}{\mathscr B}
\newcommand{\F}{\mathscr F}
\newcommand{\G}{\mathscr G}
\newcommand{\V}{\mathscr V}
\newcommand{\I}{\mathscr I}
\newcommand{\D}{\mathscr D}
$In order to better discuss this problem it is useful to have the precise categorical definition of a natural transformation in mind.
So, suppose that $\A$ and $\B$ are categories, and that $\F$ and $\G$ are functors from $\A$ to $\B$.
Definition.  A natural transformation from $\F$ to $\G$ is a correspondence that, to each object $X$ in $\A$,
assigns a morphism
$$
  \Phi_X: \F(X)\to \G(X),
  $$
satisfying the following crucial property:  if $X$ and $Y$ are objects in $\A$, and if  $f\in \text{Hom}(X, Y)$, then the
diagram
$\require{AMScd}$
\begin{CD}
  \mathscr{F}(X) @>{\Phi_X}>> \mathscr{G}(X)\\
  @V\mathscr{F}(f)VV @VV\mathscr{G}(f)V\\
  \mathscr{F}(Y) @>{\Phi_Y}>> \mathscr{G}(Y)
  \end{CD}
commutes.
This is, of course, presuming that both $\F$ and $\G$ are covariant functors.  If, on the other hand, both $\F$ and $\G$
are contravariant then the above diagram must be replaced by
\begin{CD}
\mathscr{F}(X) @>{\Phi_X}>> \mathscr{G}(X)\\
@A\mathscr{F}(f)AA @AA\mathscr{G}(f)A\\
\mathscr{F}(Y) @>{\Phi_Y}>> \mathscr{G}(Y)
\end{CD}
The question as to whether  there is a natural isomorphism from a vector space to its dual,  if phrased in the context
of category theory, should therefore be:
Question.  Considering the category $\V$ of vector spaces over a given field,  is there a natural transformation $\Phi$
from the identity functor
$$
  \I: \V\to \V,
  $$
to the dual functor
$$
  \D: \V\to \V,
  $$
namely the functor sending a vector space to its dual,  and such that $\Phi_X:X\to X^*$ is an isomorphism for every vector
space $X$?
The first major difficulty one encounters is that the identity functor $\I$ is covariant, while $\D$ is contravariant, so one
should first attempt to adapt the notion of a natural transformation for this situation.  Although this will soon be
understood  to be  a somewhat
foolish  pursuit, in the general case where our functor $\F$ above is covariant and $\G$, contravariant, we could ask
for the commutativity of the diagram
\begin{CD}
\mathscr{F}(X) @>{\Phi_X}>> \mathscr{G}(X)\\
@V\mathscr{F}(f)VV @AA\mathscr{G}(f)A\\
\mathscr{F}(Y) @>{\Phi_Y}>> \mathscr{G}(Y)
\end{CD}
Applying  this to the functors $\I$ and $\D$,  we would be asking for an a priori choice of isomorphism
$$
  \Phi_X:X\to X^*,
  $$
for every vector space $X$, such that
\begin{CD}
X @>{\Phi_X}>> X^*\\
@VTVV @AAT^*A\\
Y @>{\Phi_Y}>> Y^*
\end{CD}
commutes for every linear map $T$ from $X$ to $Y$.
However this is impossible for many reasons,  e.g. we may apply it twice,  once for $T$ and another time for $2T$,
leading up to
$$
  \Phi_X = T^*\Phi_YT = (2T)^*\Phi_Y(2T) = 4 T^*\Phi_YT = 4\Phi_X.
  $$
Of course this cannot hold!
