= Solution
If $\alpha^*=g(\alpha)$, then $\alpha$ commutes with its adjoint because it commutes with every <polynomial> in itself. Hence $\alpha$ is a <normal operator>.
Conversely let $\lambda_1,\ldots,\lambda_r$ be the distinct <eigenvalues> of a normal $\alpha$. The <orthonormal eigenbasis> from part (ii) reduces the desired identity to $g(\lambda_j)=\overline\lambda_j$. <Lagrange interpolation polynomial> provides the polynomial
$$
\boxed{g(z)=\sum_{j=1}^{r}\overline\lambda_j
\prod_{\ell\ne j}\frac{z-\lambda_\ell}{\lambda_j-\lambda_\ell},
\qquad \alpha^*=g(\alpha).}
$$
Repeated eigenvalues cause no difficulty because interpolation uses only distinct values. If $V=\{0\}$, use $g=0$. This proves both directions of the third equivalence.
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