Call a smooth function flat (in zero) if $f^{(k)}(0) = 0$ for all $k\ge 1$. Does there exist a flat Schwartz function $f\in\mathcal S(\mathbb R)$ such that its Fourier transform $\hat f = \mathcal F(f)$ is flat, too?
The construction of such a function has proven difficult to me. The simplest example of flat Schwartz functions are Bump functions which are constant around zero. However since, by the Paley-Wiener theorem, the Fourier transform of a bump function is real analytic, it cannot be flat as the Taylor series in zero is a constant.
Also note that the important identity
$$ \int_{-\infty}^{+\infty} f(x) x^k dx = \mu_k \hat f^{(k)}(0) $$
implies that all but the zeroth moment of the fourier transform of a flat function vanish. Typically the graph of a function whose moments are almost all zero looks similar to a sinc function: oscillating indefinitely and decaying as $|x|\to\infty$.
As a second remark, if such a function exists, it can w.l.o.g. be chosen as an Eigenfunction of the Fourier transform since for arbitrary $f$
$$ g = f + \mathcal F(f) + \mathcal F^2(f) + \mathcal F^3(f) $$
is an Eigenfunction of the Fourier transform, and the property of the derivatives being $0$ in the origin carries over to $g$ (since $\mathcal F^2(f)(t) = f(-t)$). This leads to a slight reformulation
Question: Does there exist an (even) Schwartz function $f\in\mathcal S(\mathbb R)$ such that
$f$ is a density function: $\int_{-\infty}^{+\infty} f(x) dx = 1$
All higher moments of $f$ vanish: $\int_{-\infty}^{+\infty} f(x)x^k dx = 0$ for all $k\ge 1$
$f$ is an Eigenfunction of the Fourier transform: $\mathcal F(f) = f$
Any ideas, remarks, or references to relevant literature are greatly appreciated!
EDIT: 7th JUL
I am still interested in this question and have started a bounty. My progress towards finding such a function has been minimal. Although it is easy to find Schwartz functions that are flat - take e.g. $e^{-x^2-1/x^2}$ - it seems very hard to impossible to control both the moments in time and frequency space simultaneously. It doesn't help that concrete examples like the above have seemingly no closed form Fourier transform. Moreover forcing $f(0)=1$ makes thing even harder. The simplest example of such a flat schwartz function I was able to find is $\exp(-\exp(x^2-1/x^2))$.
On the other hand there might be an avenue towards disproving the existence of such function. As pointed out above a flat function cannot be real analytic, but there are theorems connecting analyticity of the Fourier transform to properties like its decay rate. However the one I am familiar with (Schwartz-Payley-Wiener) deals with functions of compact support only. There seem to be generalizations with some interesting discussion found here https://mathoverflow.net/questions/23679/fourier-transform-of-analytic-functions