Duality Property for DFT
Duality in DFT would mean that if $x[n]$ has DFT coefficients as $X[k]$, then DFT of $X[n]$ would be $Nx[(N-k) \mod N]$

Proof:
Given, $$X[k] = \sum^{N-1}_{n=0}x[n]e^{-j\frac{2\pi}{N}nk}, k=0,1,2,3,...,(N-1)$$
If we take DFT of the sequence $X[n]$, then what we get is the following :
$$Y[k] = \sum^{N-1}_{n=0}X[n]e^{-j\frac{2\pi}{N}nk} = N \left(\frac{1}{N}\sum^{N-1}_{n=0}X[n]e^{j\frac{2\pi}{N}n(-k)} \right)$$
Notice that the expression between &quot;()&quot; is the synthesis expression with $k^{th}$ frequency index replaced by $(N-k \mod N)^{th}$ index. Because, in DFT expression, $k$ can only take indices $0,1,2,3,...,(N-1)$, hence, we cannot have $-k$ as frequency index, but rather, $((N-k) \mod N)$
$$Y[k] = N \left(\frac{1}{N}\sum^{N-1}_{n=0}X[n]e^{j\frac{2\pi}{N}n(N-k)} \right) = Nx[(N-k) \mod N]$$
So, the way to interpret this is, you get a scaled and inverted sequence back when you take DFT of DFT, but $x[0]$ remains at $0^{th}$ index.
So, DFT of DFT of time-domain $x[n]$ gives $\{Nx[0], Nx[N-1], Nx[N-2], ..., Nx[2], Nx[1]\}$
Scaling by $N$ is the consequence of not dividing by $\frac{1}{\sqrt{N}}$ when taking DFT and incorporating this factor into IDFT expression.