# Flip operator's equality to Double Fourier

## Main Question or Discussion Point

Some authors say that applying Fourier transform twice flips the vector, F2[x(t)] = x(-t). Yet, the simple checks proves this wrong. For instance, take 2x2 DFT:

$$\left[\begin{array}{cc}1&1\\ 1&-1\end{array}\right]^2 = \left[\begin{array}{cc}1&0\\ 0&1\end{array}\right]$$

The Identity is different from counter-identity. It cannot therefore flip the vector elements. Where is the mistake?