TITLE: Can we say that $V \cong V^*$ is not natural?
QUESTION [9 upvotes]: Can we say that an isomorphism $V \cong V^*$ is not natural? It seems so intuitively, but formally the notion of a natural transformation between a functor and a cofunctor is not defined (or is it?).

REPLY [19 votes]: This is more of a long comment on wildildildlife's answer. A natural question after reading that proof is: what happens if we restrict to the groupoid of finite dimensional vector spaces and their isomorphisms? (It may seem that picking $L=0$ is a bit of a "cheat".) The fact is that there is still no equivalence between the identity and the dual functor. The point is that the collection of isomorphisms $\{\alpha_V \colon V \to V^\ast \}$ is only allowed to depend on $V$, i.e. there must be a specific morphism for every $V$ that makes every diagram you can cook up using your category commutative. 
In particular we can look at diagrams of the form
$
\begin{array}{ccc}
 V & \longrightarrow^{\alpha_V} & V^*  \\
 \downarrow & & \uparrow  \\
 V & \longrightarrow^{\alpha_V} & V^*. 
\end{array}$
Now we get $\alpha_V = A^\ast \alpha_V A$ for all $A \in \mathrm{GL}(V)$. Equivalently, if we use $\alpha_V$ to identify two bases of $V$ and $V^\ast$, we must have $\mathrm{id}_V = A^T A$ for all $A \in \mathrm{GL}(V)$, which is absurd.
This point of view also leads directly to what condition we need to impose on the category for the existence of an equivalence like this: we need to find a category of vector spaces where the automorphisms $A \colon V \to V$ are the orthogonal matrices (wrt a basis). The most natural thing to consider is then finite dimensional spaces equipped with an inner product1 and morphisms preserving the inner product. Hence we have rediscovered the principle that you can identify a vector space with its dual precisely when it has an inner product.  
1 Remark: By an inner product I really mean an arbitrary nondegenerate bilinear form, so over $\mathbf R$ or $\mathbf C$ this is non-standard usage. Note that this category is automatically a groupoid, since any map of vector spaces preserving the inner product is an isomorphism.