Given two $n\times n$ matrices $A(z), B(z)$ with entire functions entries; with
$A$ and $B$ invertible on $\mathbb{C}$, and $A$ is normal (i.e., $A(z)A^{*}(z)=A^{*}(z)A(z)$ on $\mathbb{C}$).
If $A(z)B(z)=B(z)A(z)$ , and $A(z)B^{*}(z)=B^{*}(z)A(z)$ for all $z\in \mathbb{C}$,
does this imply that
$$A(z)B(w)=B(w)A(z) , \quad \text{for all}\; z,w\in \mathbb{C}?$$
Added: May be showing if $A(z)A(w)=A(w)A(z)$ could help!
Let $$A(z)=B(z)= \begin{bmatrix} z & z-1 \\ z+1 & z \end{bmatrix}.$$ Then $A$ is invertible and normal for all $z$. However, $A(1)B(2)\neq B(2)A(1)$.