The unitary reduction of normal matrices is a well-known fact: if $A\in M_n(\mathbb C)$ commutes with its Hermitian adjoint $A^*$, then there exists a unitary $U\in\mathbb U_n$ and a diagonal matrix $D$ such that $A=U^*DU$. And conversely, such a $U^*DU$ is normal.

Besides, the theorem of Amitsur & Levitski tells us that the standard polynomial $\mathcal S_{2n}$ in non-commutative variables vanishes identically over $M_n(k)$, where $$\mathcal S_p(A_1,\ldots,A_p):=\sum_\sigma \epsilon(\sigma)A_{\sigma(1)}\cdots A_{\sigma(p)},$$ where $\sigma$ runs over the symmetric group $\mathfrak S_p$.

Now, let me say that a matrix $A$ is $q$-normal if $\mathcal S_{2q}$ vanishes identically over the sub-algebra spanned by $A$ and $A^*$. For instance, a $1$-normal matrix is normal, whereas every $n\times n$ matrix is $2n$-normal.

Given a $q$-normal matrix $A$. Does there exist a unitary matrix $U$ and a block-diagonal matrix $D$ with diagonal blocks of sizes $\le q$, such that $A=U^*DU$ ?