Suppose the Fourier transform of $x(t)$ is $X(f)$ where 
$$X(f) = \int_{-\infty}^{\infty} x(t) \exp(-i2\pi f t) \mathrm dt$$
where $i = \sqrt{-1}$.  The inverse transform is
$$x(t) = \int_{-\infty}^{\infty} X(f) \exp(i2\pi f t) \mathrm df$$

Some properties of the Fourier transform are as follows:

 - The Fourier transform of $X(t)$ is $x(-f)$

 - If $x(t)$ is a real-valued even function of $t$, then $X(f)$
is a real-valued even function of $f$.

Thus, if $x(t)$ is a real-valued even function of $t$, then
the Fourier transform of the real-valued even function 
$X(t)$ is $x(f)$

Now suppose that $x(t)$ is an even probability density function
(so that $x(t) \geq 0$ for all $t$) with the additional property 
that $x(0) = 1$. Suppose also that its Fourier transform $X(f)$ 
has the property that $X(f) \geq 0$ for all $f$.  Then, since
$$x(0) = 1 = \int_{-\infty}^{\infty} X(f) \mathrm df$$
$X(f)$ is a even non-negative real-valued function of $f$ with
area $1$, that is, $X(f)$ is also a probability density function
with the property that $X(0) = 1$. One example of such a pair of
functions is the normal distribution cited by OP Neil G
$$x_1(t) = \exp(-\pi t^2), ~~ X_1(f) = \exp(-\pi f^2)$$
and another example is
$$x_2(t) = (1 - |t|)\mathbf 1_{[-1,1]}, ~~ 
X_2(f) = \frac{(\sin(\pi f))^2}{(\pi f)^2}.$$

Now note that  $\frac{1}{2} x_2(t) + \frac{1}{2}X_2(t)$
is a _mixture_ density whose Fourier transform is
$\frac{1}{2} X_2(f) + \frac{1}{2}x_2(f)$ which is 
_the same mixture density._

Thus, if $x(t)$ is a even density function with
$x(0) = 1$, then its
Fourier transform $X(f)$ is an even real-valued
function, and if $X(f) \geq 0$ for all $f$, then
$X(f)$ is also a density function.  From this
pair of functions we construct the density
function $\frac{1}{2} x(t) + \frac{1}{2}X(t)$
which is its own Fourier transform.

Finally, given two densities that are their own
Fourier transforms, e.g. $x_1(t)$ and 
$\frac{1}{2} x_2(t) + \frac{1}{2}X_2(t)$,
_any_ mixture density
$$\alpha x_1(t) + (1-\alpha)[\frac{1}{2} x_2(t) + \frac{1}{2}X_2(t)]$$
where $\alpha \in [0,1]$ is a density function that is its
own Fourier transform.