Let $n$ be a positive integer. The $n$ by $n$ Fourier matrix may be defined as follows:

$$ F^{*} = (1/\sqrt{n}) (w^{(i-1)(j-1)}) $$

where

$$ w = e^{2 i \pi /n} $$

is the complex $n$-th root of unity with smaller positive argument and $*$ means transpose -conjugate.

It is well known that $F$ is diagonalizable with eigenvalues $1,-1,i,-i$

where $i^2 =-1.$

It is also known that $F$ has real eigenvectors:

COMMENT: (I was unable to got this paper)

McClellan, James H.; Parks, Thomas W. Eigenvalue and eigenvector decomposition of the discrete Fourier transform. IEEE Trans. Audio Electroacoust. AU-20 (1972), no. 1, 66--74. END of COMMENT

QUESTION:

There is some simple manner to get just one of these real eigenvectors.

For example how to get a real vector with an odd number $n=2k+1$ of coordinates and such that

$$ F(x) =x. $$