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Carlo Beenakker
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Given a function $f$ and its Fourier transform $\hat{f}$, then the function

$$F=f^2+\hat{f}\star\hat{f},$$

with $\star$ the convolution, is its own Fourier transform. If we require that $f$ is a probability density (square-integrable and positive semi-definite), then any $f$ with ${\cal F}f=f$ is of this form, see A. Nosratinia, Self-characteristic distributions.

Carlo Beenakker
  • 188.1k
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  • 448
  • 651