Just wondered what this might be and the result is pretty interesting at least in the discrete case. Here it goes:

Consider a funciton ${f}$ defined on ${\Omega = \{0,1,2,\cdots,N-1\}}$. The discrete Fourier transform is given by: ${\hat{f}(\omega) = \sum_{k=0}^{N-1}f(k)e^{\frac{-2\pi ik\omega}{N}} \forall \omega \in \Omega}$

The discrete Fourier transform of the discrete Fourier transform is given by

$\displaystyle \begin{array}{rcl} \hat{\hat{f}}(\eta) &=& \sum_{l=0}^{N-1}\hat{f}(l) e^{\frac{-2\pi il\eta}{N}}\\ &=&\sum_{l=0}^{N-1}\left(\sum_{k=0}^{N-1}f(k)e^{\frac{-2\pi ikl}{N}}\right) e^{\frac{-2\pi il\eta}{N}}\\ &=&\sum_{k=0}^{N-1}\left(f(k)\sum_{l=0}^{N-1}e^{\frac{-2\pi il(k+\eta)}{N}} \right)\forall \eta \in \Omega \end{array}$

Now, the sum ${ S(k,\eta) = \sum_{l=0}^{N-1}e^{\frac{-2\pi il(k+\eta)}{N}} \forall k \in \Omega, \forall \eta \in \Omega}$ is special. Doing the usual geometric progression thing:

$\displaystyle \begin{array}{rcl} S(k,\eta)(1 - e^{\frac{-2\pi i(k+\eta)}{N}}) &=& 1 - e^{\frac{-2\pi iN(k+\eta)}{N}} \\ &=& 0 \end{array}$

So, if ${k+\eta}$ is not a multiple of ${N}$ or ${0}$ then ${S(k,\eta) = 0}$ for that pair of ${k,\eta}$. When ${\eta = 0}$, only for ${k=0}$, ${1 - e^{\frac{-2\pi iN(k+\eta)}{N}} = 0}$ and this value can be easily computed to be ${Nf(0)}$. Therefore, ${\hat{\hat{f}}(0) = Nf(0)}$ When ${\eta=1}$, only ${k=N-1}$ makes ${1 - e^{\frac{-2\pi iN(k+\eta)}{N}} = 0}$ and only this term would remain in the sum. A simple computation yields ${\hat{\hat{f}}(1) = Nf(N-1)}$. Simlarly it can be shown that ${\hat{\hat{f}}(\eta) = Nf(N-\eta), \forall \eta \in \Omega \backslash \{0\}}$. Putting it all together:

${\hat{\hat{f}}(\eta) = \left\{ \begin{array}{cc} Nf(0);&\mathrm{if} \; \eta = 0,\\ Nf(N-\eta);& \forall \eta \in \Omega \backslash \{0\} \end{array} \right. }$

What is interesting is that to compute the values of the original function one can also do a forward transform (instead of the inverse transform).

Upon further thought I found that there is a much simpler one-line proof:

$\displaystyle \begin{array}{rcl} Nf(\eta) &=& \sum_{l=0}^{N-1}\hat{f}(l) e^{\frac{2\pi il\eta}{N}}\\ &=& \hat{\hat{f}}(N - \eta), \forall \eta \in \Omega \backslash \{0\} \end{array}$

the case of ${\eta=0}$ being trivial. So, ofcourse the forward and backward transforms of the fourier transform are related.