Suppose we have a family of isomorphisms $\alpha_A\colon A\to A^*$ for all finite abelian groups $A^*$. If $\phi\colon A\to B$ is an isomorphism, we have a dual isomorphism $\phi^*\colon B^*\to A^*$ and thus an isomorphism $(\phi^*)^{-1}\colon A^*\to B^*$. This makes $A^*$ a covariant functor of $A$ on the category of finite abelian groups and isomorphisms, and it is only reasonable to call $\alpha$ canonical if it is natural with respect to this structure. In other words, we should have $(\phi^*)^{-1}\circ\alpha_A=\alpha_B\circ\phi$ for all $\phi$, or $\alpha_A=\phi^*\circ\alpha_B\circ\phi$. In particular, this must hold when $B=A$ and $\phi=n.1_A$ for some $n$ that is coprime to the order of $A$. This means that $(n^2-1).\alpha_A=0$, but $\alpha_A$ is assumed to be an isomorphism, so the exponent of $A$ must divide $n^2-1$. This fails when $A=\mathbb{Z}/5$ and $n=2$, for example, so there is no natural map $\alpha$ as described.

The same argument also shows that there is no natural isomorphism $V\to V^*$ for finite-dimensional vector spaces $V$ over $\mathbb{Z}/p$ provided that $p>3$. The same conclusion holds for $p=2$ or $p=3$ but one needs to use some different choices of $\phi$ to prove it.

On the other hand, if we restrict even further to elementary abelian groups of order $4$, then you can check that there is a natural choice of $\alpha_A$. It sends each nonzero element $a\in A$ to the unique map $\theta\colon A\to\mathbb{Z}/2$ such that $\theta\neq 0$ but $\theta(a)=0$.

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