When people say 'canonical' what they mean in this context is something like 'definable without parameters' (i.e., without choosing bases; actually, without choosing anything at all.) See for instance the entry for "Definable Set" in wikipedia. The important point is that canonical objects are invariant under automorphisms that preserve the relevant structure.

This means our isomorphism $V\to V^*$ should be invariant under all automorphisms of k-Vect that preserve the dualizing functor, considered as a contravariant functor from k-Vect to k-Vect. This is not possible to obtain even for one finite-dimensional $V$. The reason is that with respect to some bases $B,C$ on $V,V^*$, a given isomorphism has the form of an identity matrix. But for any matrix $A$ representing an isomorphism $V\to V$, there is an automorphism of k-Vect which preserves the dual space functor and which transforms the representation of the given automorphism (with respect to the same bases $B,C$) to $A^T A^{-1}$. We can find a matrix $A$ for which this expression is different than $I$ (unless $\dim(V)=1$ or $|k|=2=\dim(V)$, as I learned here). Thus the isomorphism $V \to V^*$ is not preserved by this automorphism.

$\newcommand\kVect{k\text{-Vect}}$When people say ‘canonical’ what they mean in this context is something like ‘definable without parameters’ (i.e., without choosing bases; actually, without choosing anything at all). See for instance the entry for “Definable Set” in Wikipedia. The important point is that canonical objects are invariant under automorphisms that preserve the relevant structure.

This means our isomorphism $V\to V^*$ should be invariant under all automorphisms of $\kVect$ that preserve the dualizing functor, considered as a contravariant functor from $\kVect$ to $\kVect$. This is not possible to obtain even for one finite-dimensional $V$. The reason is that with respect to some bases $B$, $C$ on $V$, $V^*$, a given isomorphism has the form of an identity matrix. But for any matrix $A$ representing an isomorphism $V\to V$, there is an automorphism of $\kVect$ which preserves the dual space functor and which transforms the representation of the given automorphism (with respect to the same bases $B$, $C$) to $A^T A^{-1}$. We can find a matrix $A$ for which this expression is different than $I$ (unless $\dim(V)=1$ or $\lvert k\rvert=2=\dim(V)$, as I learned here). Thus the isomorphism $V \to V^*$ is not preserved by this automorphism.

When people say 'canonical' what they mean in this context is something like 'definable without parameters' (i.e., without choosing bases.) See for instance the entry for "Definable Set" in wikipedia. The important point is that canonical objects are invariant under automorphisms that preserve the relevant structure.

This means our isomorphism $V\to V^*$ should be invariant under all automorphisms of k-Vect that preserve the dualizing functor, considered as a contravariant functor from k-Vect to k-Vect. This is not possible to obtain even for one finite-dimensional $V$. The reason is that with respect to some bases $B,C$ on $V,V^*$, a given isomorphism has the form of an identity matrix. But for any matrix $A$ representing an isomorphism $V\to V$, there is an automorphism of k-Vect which preserves the dual space functor and which transforms the representation of the given automorphism (with respect to the same bases $B,C$) to $A^T A^{-1}$. We can find a matrix $A$ for which this expression is different than $I$ (unless $2=\operatorname{char}(k)=\dim(V)$, as I learned here). Thus the isomorphism $V \to V^*$ is not preserved by this automorphism.

When people say 'canonical' what they mean in this context is something like 'definable without parameters' (i.e., without choosing bases; actually, without choosing anything at all.) See for instance the entry for "Definable Set" in wikipedia. The important point is that canonical objects are invariant under automorphisms that preserve the relevant structure.

This means our isomorphism $V\to V^*$ should be invariant under all automorphisms of k-Vect that preserve the dualizing functor, considered as a contravariant functor from k-Vect to k-Vect. This is not possible to obtain even for one finite-dimensional $V$. The reason is that with respect to some bases $B,C$ on $V,V^*$, a given isomorphism has the form of an identity matrix. But for any matrix $A$ representing an isomorphism $V\to V$, there is an automorphism of k-Vect which preserves the dual space functor and which transforms the representation of the given automorphism (with respect to the same bases $B,C$) to $A^T A^{-1}$. We can find a matrix $A$ for which this expression is different than $I$ (unless $\dim(V)=1$ or $|k|=2=\dim(V)$, as I learned here). Thus the isomorphism $V \to V^*$ is not preserved by this automorphism.