Is the identity functor naturally isomorphic to a covariant dual functor? It is often said that vector spaces are not naturally isomorphic to dual spaces, because the dual functor is not naturally isomorphic to the identity functor.  But the latter is a rather trivial statement, because you can’t have a natural transformation between a contravariant functor and a covariant functor.  So I’d like to see if it it can be modified into something less trivial.
Let $VectIso$ be the category with vector spaces as objects and bijective linear transformations as morphisms.  Let $D:VectIso\rightarrow VectIso$ be a covariant functor defined by $D(V)=V^*$ for any vector space $V$ and $D(T)(f)=f\circ T^{-1}$ for any bijective linear transformation $T:V\rightarrow W$ and any linear functional $f\in V^*$.
Then my question is, is $D$ naturally isomorphic to the identity functor?  I assume the answer is no, but how would you prove it?
 A: Suppose we had a natural transformation $\eta: 1 \Rightarrow D$.  Let $V$ be a vector space and $T \in GL(V)$.  The naturality condition asserts that for all $v \in V$, we have $$\eta_V(Tv) = \eta_V(v) \circ T^{-1}.$$  Fix $v \in V \setminus \{0\}$, and let $f = \eta_V(v)$.  Since $GL(V)$ acts transitively on $V \setminus \{0\}$, we see that $\eta_V: V \to V^*$ is completely determined by $f$ according to the equation above.  
We need to check that $\eta_V$ is thus well-defined.  In particular, we need to have $$f = f \circ T$$ for all $T$ fixing $v$.  
Let us be more concrete.  Let $V = \mathbb{R} \langle e_1, e_2, e_3 \rangle$ and $v = e_1$.  Consider transformations $T \in GL(V)$ as follows: $$\begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \qquad \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}, \qquad \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}.$$  These matrices all fix $e_1$.  Plugging them into the equation $f = f \circ T$, we find
\begin{align}
f(e_2) &= f(e_1) + f(e_2) \\
f(e_3) &= f(e_2) + f(e_3) \\
f(e_2) &= f(e_3)
\end{align}
Consequently, we have $f(e_1) = f(e_2) = f(e_3) = 0$, so that $f$ is the zero linear functional.  But then $\eta_V(e_1) = 0 = \eta_V(0)$, contradicting the fact that $\eta_V$ is a linear isomorphism.  So no natural transformation $\eta: 1 \Rightarrow D$ can exist.  
A: First note that you should require only to work with finite dimensional vector spaces, otherwise the answer is no for trivial reasons (i.e. that a vector space need not be isomorphic to its dual). 
With this in mind, the answer (somewhat unsurprisingly) stays no. For this I'll use the well-known equivalence between the category of finite dimensional vector spaces and the category of integers, with matrices as arrows. Then your category is the core of the category of finite dimensional vector spaces, so it is equivalent to the core of the category of integers with matrices, that is, to the coproduct of the groups $GL(n, k)$ (seen as categories with one object). 
On this category, the dual functor becomes the identity on objects, and the transpose of the inverse on arrows : $f\mapsto (f^{-1})^T$ (here I'm cheating a bit, because I assumed that the equivalence between your category and its skeleton was compatible with duals, that is if $V$ is sent to $n$ with basis $(e_i)$, then $V^*$ should be given the dual basis : but that can be solved, because a dual can be identified set-theoretically, i.e. $V$ can be recovered from $V^*$ and the rank of $V$ - ordinal rank - is strictly smaller than that of $V^*$ so you can make your choices consistently; that's a bit of set-theoretic trickery)
Thus a natural isomorphism between that and the identity amount sto a family $(\alpha_n)$ of elements of $GL(n,k)$ such that for all $f\in GL(n,k), (f^{-1})^T=\alpha_n\circ f\circ \alpha_n^{-1}$. 
Taking for instance for $f$ a permutation matrix yields that $\alpha_n$ commutes with those; and a quick computation indicates that this means sonething like the result here; now taking $f$diagonal with entries either $1$ or $-1$ gives you that the nondiagonal entries must be $0$ (here I'm assuming that $1\neq -1$ in $k$ - or by taking different diaginal entries, that the field has at least $3$ elements); so $\alpha_n$ is a scalar, which yields $(f^{-1})^T=f$ for all $f$ which is of course absurd for any field $k$
(If you know enough linear algebra, you can be quicker : this would imply that $f$ and $f^{-1}$ are always conjugate, in particular they would have the same eigenvalues, which is of course absurd for any field, for $n$ large enough)
