Suppose that $T\in \mathcal L (V)$. prove that T is normal iff there exists a polynomal $f(x) \in \Bbb C [x] $ s.t $T^* = f(T)$
$(\Leftarrow) $ notice that $T^* T= f(T) T = T f(T) = T T^* $ as T commutes with a polynomail in T.
$(\Rightarrow ) $ Let $ v_1 , \cdots , v_n $ be an orthonormal basis for T.
$$ p(x) = \sum_{i=0}^{k} a_i x^{i} $$ $$ p(T) = a_0 I + a_1 T + \cdots + a_k T{k} $$ $$p(T)v_j = a_0 v_j + a_1 \lambda_{i} v_j + a_2 \lambda_{i}^2 v_j \cdots + a_k \lambda_{i}^{k} v_j $$ $$ = p(\lambda_i) v_j $$ $$ =\bar {\lambda_{i}} v_{j} $$ $$= T^{*} v_j $$
If $T$ is normal, then we have $Tv = \lambda v$ if and only if $T^*v = \overline{\lambda}v$.
If $\sigma(T) = \{\lambda_1, \ldots, \lambda_n\}$, let $p$ be a polynomial such that $p(\lambda_i) = \overline{\lambda_i}$ for all $i = 1,\ldots,n$.
If $Tv =\lambda v$ we have:
$$p(T)v = p(\lambda) v = \overline{\lambda} v = T^*v$$
Since $T$ diagonalizes, we can decompose $x\in V$ as $x = v_1 + \cdots + v_n$ where $v_i \in \ker (T - \lambda_i I) = \ker (T^* - \overline{\lambda_i} I)$.
Now we have
$$p(T)x = p(T)v_1 + \cdots + p(T)v_n = T^*v_1 + \cdots + T^*v_n = T^{*}x$$
Therefore $T^* = p(T)$.