Which function's Fourier transform is the function itself?

Pick a function $f$ that is reasonable enough for the inversion formula to hold (e.g. take $f$ in Schwartz space, which contains the Gaussian among other functions). If $\mathcal{F}$ denotes the linear transformation which takes $f$ to its Fourier transform, then it's easy to check that $\mathcal{F}^{4}$ is the identity map. In particular, by playing some games you find that $$g \ = \ f + \mathcal{F}(f) + \mathcal{F}^{2}(f) + \mathcal{F}^{3}(f) $$ is fixed by $\mathcal{F}$. So $g$ is its own Fourier transform.

This argument doesn't produce a concrete function, but it at least shows you that the Gaussian is far from the only function that is equal to its own Fourier transform. If you want a more specific example, you can show that $(\cosh \pi x)^{-1}$ is its own Fourier transform (use contour integration and the residue theorem).

Well, the Dirac delta function, strictly speaking, is not a 'function' but it serves your purpose.

The Fourier transform of a Dirac comb (or an impulse train, as it is called in Electrical Engineering) is another Dirac comb.

$$ \sum_{n=-\infty}^{\infty} \delta(t-nT_o) = \frac {2\pi} {T_o} \sum_{m=-\infty}^{\infty} \delta(\omega - m\omega_o) $$

For the proof, refer http://nptel.ac.in/courses/IIT-MADRAS/Principles_Of_Communication/pdf/Lecture07_FTPeriodic.pdf