From simulation, I found out that following condition almost holds with equality $\operatorname{diag}(\operatorname{diag}(AB)(\operatorname{diag}(AB))') \approx 1/T \operatorname{diag}(ABB'A') $, where complex matrices $A\in \mathcal{C}^{T\times S}$ and $B\in \mathcal{C}^{S\times T}$ are assumed with slight abuse of the notation of operator "$\operatorname{diag}$". $A$ and $B$ have complex entries of form $\frac{1}{\sqrt{S}}e^{-j\phi_{t,s}}$ or $\frac{1}{\sqrt{S}}e^{-j\theta_{s,t}}$, respectively.
So far, I can show applying the results from https://epubs.siam.org/doi/pdf/10.1137/S0895479896309645 and rewriting the "$\operatorname{diag}$" using the Hadamard product "$\cdot$" that following is fulfilled $( (AB)\cdot I^{S\times S} )( (AB)\cdot I^{S\times S} )'\le (A(BB')A')\cdot (I^{S\times S}I^{S\times S})$.
How can I introduce the scaling factor $1/T$, where $T$ is the dimension of the matrix $B$ into the equations? Can it be true that my matrices $A$ and $B$ have some further special structure I am not aware of?
EDIT: Changed the title according to the first comment. Furthermore "'" is the conjugate transpose operation, like "^H".
Define $$E = {\rm Diag}(e),\quad x={\rm diag}(X),\quad y={\rm diag}(Y)$$ Then here are some Hadamard-Diag equalities $$\eqalign{ {\rm diag}(XEY) &= (Y^T\odot X)\,e \\ {\rm diag}(X\!\odot\!Y) &= x\odot y \;=\; {\rm diag}(xy^T) \\ }$$ Setting $e={\tt1}$ (the all ones vector), then $E=I$ (the identity matrix) and $$\eqalign{ {\rm diag}(XY) &= (Y^T\odot X)\,{\tt1} \\ }$$ Applied to the current problem $$\eqalign{ {\rm diag}\Big({\rm diag}(AB)\;{\rm diag}(AB)^T\Big) &= {\rm diag}(AB) \odot {\rm diag}(AB) \\ &= \big((B^T\odot A){\tt1}_s\big)\odot\big((B^T\odot A){\tt1}_s\big) \\ {\rm diag}(BAA^TB) &= (B^TA\odot BA){\tt1}_s \\ }$$ This is an incomplete analysis which does not consider the special form of the elements of $A$ and $B$. Or whether the matrices are symmetric or hermitian.