The Fourier transform $\hat u$ is defined on the Schwartz space $\mathscr S(\mathbb R^n)$
by
$
\hat u(\xi)=\int e^{-2iπ x\cdot \xi} u(x) dx.
$
It is an isomorphism of $\mathscr S(\mathbb R^n)$ and the inversion formula is
$
u(x)=\int e^{2iπ x\cdot \xi} \hat u(\xi) d\xi.
$
The Fourier transformation can be extended to the tempered distributions
$\mathscr S'(\mathbb R^n)$ with the formula
$$
\langle \hat T,\phi\rangle_{\mathscr S'(\mathbb R^n), \mathscr S(\mathbb R^n)}
=
\langle T, \hat \phi\rangle_{\mathscr S'(\mathbb R^n), \mathscr S(\mathbb R^n)}.
$$
Below we note $\mathcal F$ the Fourier transformation on $\mathscr S'(\mathbb R^n)$. We find easily that $\mathcal F^4=Id$, so that if $T\in\mathscr S'(\mathbb R^n)$ is such that
$\mathcal F T=\lambda T$, then $\lambda$ is a fourth root of unity.
Question: Are all the tempered distributions $T$ such that $\mathcal FT=T$ known? Two examples are very classical: first the Gaussians $e^{-π\vert x\vert^2},
e^{-π\langle Ax,x\rangle } $ where $A$ is a positive definite matrix with determinant 1, second the case
$$
T_0=\sum_{k\in \mathbb Z^n}\delta_k,
$$
where the equality $\mathcal FT_0=T_0$ is the Poisson summation formula.
I think that, thanks to your answers and a reference, I found the answer: using What are fixed points of the Fourier Transform which deals with the $L^2$ case, one may guess that the fixed points of $\mathcal F$ in $\mathscr S'(\mathbb R^n)$ are
$$
\Bigl\{S+\mathcal F S+\mathcal F^2 S+\mathcal F^3 S
\Bigr\}_{S\in \mathscr S'(\mathbb R^n)}=(Id+\mathcal F+\mathcal F^2+\mathcal F^3)(\mathscr S'(\mathbb R^n)).
$$
Since $\mathcal F^4=Id$ on $\mathscr S'(\mathbb R^n)$, the above distributions are indeed fixed points and if $\mathcal F T=T$, then
$$
T=\frac14\bigl(T+\mathcal F T+\mathcal F^2 T+\mathcal F^3 T\bigr),
$$
conluding the proof.
Best Answer
Given a square-integrable, positive semi-definite function $f$, with 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: $\hat{F}=F$.
If we require that $F$ is a probability density (absolutely integrable and positive semi-definite), then any $F$ with $\hat{F}=F$ is of this form, see A. Nosratinia, Self-characteristic distributions.
The decomposition $F=f^2+\hat{f}\star\hat{f}$ for a given probability density $F=\hat{F}$ is not unique, one realization is $f=\sqrt{F/2}$.