DFT's Unitary Property
Just so we're on the same page: given some sequence $x[n]$ for $n=0, \dots, N-1$, the $k$'th DFT coefficient is (forgetting the scaling factor)

$$X[k] = \sum_{n=0}^{N-1} x[n] \exp(-j 2\pi k n/N)$$ 

We can get all the coefficients by writing this as $ X = Dx$, where $D$ is the $N \times N$ DFT matrix and $D_{kn} = e^{-j 2\pi k n /N}$. For the N = 2 case, 

$$D = \begin{pmatrix} 
e^{-j 2\pi (0)(0) /2} &amp;  e^{-j 2\pi (0)(1)/2}\\
e^{-j 2\pi (1)(0) /2} &amp;  e^{-j 2\pi (1)(1)/2}
\end{pmatrix} = 
\begin{pmatrix} 
e^{-j 0} &amp;  e^{-j 0}\\
e^{-j 0} &amp;  e^{-j \pi}
\end{pmatrix} = \begin{pmatrix} 
1 &amp;  1\\
1 &amp;  -1
\end{pmatrix},$$
for which $D^* D = D^T D = D^2 = 2I$, so (minus the scaling factor), unitary.