This$\DeclareMathOperator\DNN{DNN}\DeclareMathOperator\CP{CP}$This question can be thought of as an offshoot of this MO question from a few months ago. Let $M_n(\mathbb{C})$ denote the set of $n \times n$ complex matrices. Whenever I refer to positive semidefinite (PSD) matrices, I assume they are Hermitian. I use "$\odot$" to refer to the entrywise/Hadamard product of matrices, and I will shortly take the (entrywise) complex conjugate of a matrix.
Vague question: What can be said about the set $S$ of matrices $$ S := \{ X \odot \overline{X} : X \in M_n(\mathbb{C}) \text{ is positive semidefinite}\}? $$$$ S \mathrel{:=} \{ X \odot \overline{X} : \text{$X \in M_n(\mathbb{C})$ is positive semidefinite}\}? $$
Some observations: Every member of $S$ is clearly real and entrywise non-negative, and the solution to the MO question linked abovethe MO question linked above shows that every member of $S$ is positive semidefinite. We thus have $S \subseteq DNN$$S \subseteq \DNN$, where $DNN$$\DNN$ is the set of doubly nonnegative matrices.
Precise question #1: Does $S = DNN$$S = \DNN$?
If the answer to "precise question #1" is "no", several natural follow-up questions arise. For example, the most well-known set of matrices that is contained within $DNN$$\DNN$ is the set $CP$$\CP$ of completely positive matrices. This leads to... …
Precise question #2: Does $S = CP$$S = \CP$?
In fact, it's not clear to me that $S \subseteq CP$$S \subseteq \CP$ or $CP \subseteq S$$\CP \subseteq S$, so either of those inclusions (or refutations of those inclusions) would be interesting as well.
