= Solution
The <dual representation> on $V^*$ is $(g\ell)(v)=\ell(g^{-1}v)$. The bilinear form $\varepsilon(z,w)=\det(z,w)$ on $\mathbb C^2$ is nondegenerate and invariant under <SU(2)>, because every group matrix has determinant one. Thus it gives an equivariant isomorphism $\mathbb C^2\to(\mathbb C^2)^*$. Taking the $n$th <symmetric power> gives an equivariant isomorphism between its $n$th symmetric power and its dual. Since homogeneous polynomials realize $V_n=\operatorname{Sym}^n((\mathbb C^2)^*)$, this proves
$$
\boxed{V_n^*\cong V_n.}
$$
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