Let $\xi \in \{0, 1\}^n$ be a vector included into $\mathbb C^n$, and use $\mathcal F : \mathbb C^n \to \mathbb C^n$ to be any natural frequency-time space conversion map.
What can we say about
$$\mathbb \xi^{\tilde \dagger} := \mathcal {Real}(\mathcal F^{-1}(\{ u^2 : u \in \mathcal F(\xi) \})$$
First, we can note this operation creates a co-natural ordering $\pi^{\tilde {\dagger}}$ of entries of $\xi$. (Those permutations $\pi$ which rearrange entries of $\xi^{\tilde \dagger}$ for $\pi^{\tilde {\dagger}}\xi^{\tilde \dagger}$ to be monotone increasing).
I have thought about this problem a bit.. Since the input is binary, the output has no imaginary component. I've figured out this much, so the real should encode the information. Going based on the definitions of the series, maybe we can figure something out algebraically:
$$\xi^{\tilde \dagger}_b = \sum_{t=0}^{n-1} (\sum_{k=0}^{n-1} \xi_k \exp(2i \pi kb/n))^2 \exp(2i \pi tb/n)$$ $$= \sum_{t=0}^{n-1} \left( \sum_{u=0}^{n-1}\sum_{v=0}^{n-1} \xi_u \xi_v\exp\left(\frac{2i \pi b}n(u+v)\right) \right) \exp(2i \pi tb/n)$$
So by squaring each coordinate of $\mathcal F(\xi)$, it's almost like now it is a function of products of two elements of $\xi$, since the exp terms are only applying rotations. So when taking the inverse map, we will end up with images of many distinct positive amplitudes.
My question: What information about $\xi$ is revealed by the set of permutations precipitated by the operation $\cdot^{\tilde \dagger}$? Which internal structure of $\xi$ is being elucidated? Thank you!
Bonus: What can be said about $\xi^{\tilde \dagger^{\tilde \dagger}}, \xi^{\tilde \dagger^{\tilde \dagger^{\tilde \dagger}}},$ and so on?