Is the discrete Gaussian kernel an eigenfunction of the DFT?
Since the DFT is representable by multiplication with the Fourier matrix, your question is equivalent to asking what the eigenvectors of the Fourier matrix are.

Actually, Wikipedia provides the answer (http://en.wikipedia.org/wiki/Discrete_Fourier_transform#Eigenvalues_and_eigenvectors).

However, since the eigenvalues ($1, -1, i, -i$) are not simple, the eigenvectors are not unique (i.e. linear combinations are also eigenvectors). Also no simple closed formula exists.

One formula for an eigenvector close to what you ask is provided by Wikipedia

$$F_m = \sum_{k = -\infty}^\infty \exp\left(-\frac{\pi\cdot(m+N\cdot k)^2}{N}\right) \quad\quad\quad m = 0,\ldots,N-1$$

So concluding, the Gaussian function itself is not an eigenvector, but an infinite sum of Gaussians. 
The infinite sum can probably be interpreted as equivalent to the discretization of the frequency and time domain when we going from the FT to the DFT. So it is not as simple, as just truncating the discrete Gaussian.