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 density function whose Fourier transform $X(f)$ is a density function, then the mixture density function $\frac{1}{2} x(t) + \frac{1}{2}X(t)$ 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 $$\alpha x_1(t) + (1-\alpha)\left[\frac{1}{2} x_2(t) + \frac{1}{2}X_2(t)\right]$$ where $\alpha \in [0,1]$ is a density function that is its own Fourier transform.

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 $$x_2(t) = (1 - |t|)\mathbf 1_{[-1,1]}, ~~ X_2(f) = \operatorname{sinc}^2(f) = \begin{cases}\displaystyle \left(\frac{\sin(\pi f)}{\pi f}\right)^2,&f\neq 0,\\ 1,&f=0.\end{cases}$$

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 density function whose Fourier transform $X(f)$ is a density function, then the mixture density function $\frac{1}{2} x(t) + \frac{1}{2}X(t)$ 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 $$\alpha x_1(t) + (1-\alpha)\left[\frac{1}{2} x_2(t) + \frac{1}{2}X_2(t)\right]$$ where $\alpha \in [0,1]$ is a density function that is its own Fourier transform.

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.

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 density function whose Fourier transform $X(f)$ is a density function, then the mixture density function $\frac{1}{2} x(t) + \frac{1}{2}X(t)$ 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 $$\alpha x_1(t) + (1-\alpha)\left[\frac{1}{2} x_2(t) + \frac{1}{2}X_2(t)\right]$$ where $\alpha \in [0,1]$ is a density function that is its own Fourier transform.