The obvious ones are 0 and $e^{x^2}$ (with annoying factors), and someone I know suggested hyperbolic secant. What other fixed points (or even eigenfunctions) of the Fourier transform are there?
The following is discussed in a little more detail on pages 337339 of Frank Jones's book "Lebesgue Integration on Euclidean Space" (and many other places as well). Normalize the Fourier transform so that it is a unitary operator $T$ on $L^2(\mathbb{R})$. One can then check that $T^4=1$. The eigenvalues are thus $1$, $i$, $1$, and $i$. For $a$ one of these eigenvalues, denote by $M_a$ the corresponding eigenspace. It turns out then that $L^2(\mathbb{R})$ is the direct sum of these $4$ eigenspaces! In fact, this is easy linear algebra. Consider $f \in L^2(\mathbb{R})$. We want to find $f_a \in M_a$ for each of the eigenvalues such that $f = f_1 + f_{1} + f_{i} + f_{i}$. Using the fact that $T^4 = 1$, we obtain the following 4 equations in 4 unknowns: $f = f_1 + f_{1} + f_{i} + f_{i}$ $T(f) = f_1  f_{1} +i f_{i} i f_{i}$ $T^2(f) = f_1 + f_{1}  f_{i}  f_{i}$ $T^3(f) = f_1  f_{1} i f_{i} +i f_{i}$ Solving these four equations yields the corresponding projection operators. As an example, for $f \in L^2(\mathbb{R})$, we get that $\frac{1}{4}(f + T(f) + T^2(f) + T^3(f))$ is a fixed point for $T$. 


Following on a little from Andy's comment, Hermite polynomials (multiplied by a Gaussian factor) give a basis of eigenvectors for the FT as an operator on $L^2({\mathbb R})$ 


A very important fixed point of the Fourier transform that isn't in $L^2$ is the Dirac comb distribution, informally $$D(x) = \sum_{n\in Z} \delta(xn),$$ or more properly, defined by its pairing on smooth functions of sufficient decay by $$\langle D, f\rangle = \sum_{n\in Z} f(n).$$ The fact that $D$ is equal to its Fourier transform is really just the Poisson summation formula. (I wrote an argument explaining why $D$ should be its own Fourier transform in an answer to another question: Truth of the Poisson summation formula) 

