I saw the answer quite late, but thanks anyway. I found a way in the meanwhile, but I should check whether it is more efficient than just doing FFT and IFFT in two different dimensions. For the interested, it is the following:
The IFFT is the same as FFT of the inverse of the array, apart from a factor. So:
http://latex.codecogs.com/gif.latex?X_j=\sum_{k=1}^Ne^{i2\pi/Njk}
Is also equal to:
http://latex.codecogs.com/gif.latex?X_j=\frac{e^{i2\pi/Nj}}{N}\sum_{k'=1}^Ne^{-i2\pi/Njk'}x_{N-k'+1}
This can be shown by a change of index k=N-k'+1 (just follow the links for the equation). If it is possible to store the rows in reverse order, then one plan for a 2-D FFT is sufficient for the original question, but then all rows should still be multiplied by a constant.