Is the canonical isomorphism to double dual unique? Let $V$ be a finite-dimensional vector space. There is a canonical isomorphism $\varphi:V\rightarrow V^{**}$.
Is $\varphi$ unique? Namely, when $\psi:V\rightarrow V^{**}$ is another isomorphism that is defined independently of the choice of basis, do $\varphi$ and $\psi$ coincide?
 A: If the space $V$ has any nontrivial automorphism $\alpha$, then the composition $\psi=\varphi\circ\alpha$ is another isomorphism that is not identical to $\varphi$. Any vector space with dimension $>1$ has a non-trivial automorphism obtained by permuting basis elements. Therefore, in these cases, the canonical isomorphism to the double-dual is not unique.
A: The answer is negative. Over field $k \neq \mathbb{F}_2$, take $c \in k \setminus \{0, 1\}$, and $\psi = c \cdot \varphi$.
A: Edit. Here is an elementary argument. 
Let $\varphi_V:V\to V^{**}$ be the canonical morphism and let $\psi_V:V\to V^{**}$ be another functorial morphism. 
Let's insist one the fact that $\varphi_V$ and $\psi_V$ depend functorially on the vector space $V$. 
There is a scalar $\lambda$ in the ground field $K$ such that $\psi_K=\lambda\varphi_K$.
We claim that we have $\psi_V=\lambda\varphi_V$ for all $V$. 
All the maps below are $K$-linear.
Given $v:K\to V$, it suffices to show $\psi_V(v(1))=\lambda\varphi_V(v(1))$. 
But we have 
$$
\psi_V(v(1))=v^{**}(\psi_K(1))=\lambda v^{**}(\varphi_K(1))=\lambda \varphi_V(v(1)).
$$ 
End of the edit.
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