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.