Let $e_1, \ldots, e_D$ be a given orthonormal basis for $\mathbb{C}^D$. We define the phase profile of a vector $u \in \mathbb{C}^D$ to be $( \text{arg} (e_1^* u), \ldots \text{arg} (e_D^* u) )$. We take $\text{arg} (0) := 0$.
Let $A: \mathbb{C}^D \rightarrow \mathbb{C}^D$ be an arbitrary positive semi-definite operator. Is it possible to find $a_1, \ldots, a_S \in \mathbb{C}^D$ that have equal phase profile such that $A = \sum_{i=1}^S a_i a_i^*$?
If $D =2$, then the answer is yes. I suspect that it is not true for larger $D$, but I am yet to find a counter example.
When $D\ge3$, the answer is ‘no’ in general.
Let $E=\pmatrix{e_1&e_2&\cdots&e_D}$. When your requirement is satisfied and if it happens that $E^\ast AE$ is entrywise nonzero, let $(\theta_1,\theta_2,\ldots,\theta_D)$ be the common profile. Then $$ \arg\big((E^\ast AE)_{ij}\big)=\theta_i-\theta_j $$ and hence the matrix $X$ defined by $$ X_{ij}=\frac{(E^\ast AE)_{ij}}{|(E^\ast AE)_{ij}|} $$ is necessarily rank-one, because the RHS is equal to $e^{\mathbf i(\theta_i-\theta_j)}$. However, it is easy to construct an example of $(E,A)$ such that $\operatorname{rank}(X)\ge2$, such as $$ E=I_3,\quad A=\pmatrix{3&-i&1\\ i&3&1\\ 1&1&3}\quad\text{and}\quad X=\pmatrix{1&-i&1\\ i&1&1\\ 1&1&1}. $$