Eigenvectors of the Fourier matrix 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
$$
F(x) =x.
$$
 A: There is full and simple description of all eigenvectors in the article 
Morton, P. On the eigenvectors of Schur's matrix. J. Number Theory, 1980, 12, 122-127 http://deepblue.lib.umich.edu/bitstream/2027.42/23371/1/0000315.pdf
A: The vector $v = (1 - \sqrt{n}, 1, 1, 1, ...)$ is an eigenvector of $F(n)$ with an eigenvalue of -1 for all $n > 2$. To see this, note that the first row of $F(n)$ is $(\frac{1}{\sqrt{n}}, \frac{1}{\sqrt{n}}, \frac{1}{\sqrt{n}}, ...)$.  From this is follows trivially that the first element of $F(n).v$ is $-1 + \sqrt{n}$.  For all of the other rows of $F(n)$, we know that for a vector of all ones, the results of the row would sum to zero, since the row contains exactly the n roots of unity.  Given this, and the fact the the first element of any row is again $\frac{1}{\sqrt{n}}$, if we were to multiply any row of $F(n)$ other than row 1, by the vector $(0,1,1,...)$, the total for a row is always $-\frac{1}{\sqrt{n}}$.  Thus, for the vector $v$, we find for any row greater that zero, multiplying the row by $v$ results in $(1 - \sqrt{n})\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n}} = -1$
A: This has a little number-theoretic content, having to do with real-valued characters modulo $n=2k+1$. For example, for $n=p$ an odd prime number, there are exactly two such functions (up to scalar multiples), the function that is $1$ for non-zero-mod-$p$ inputs, and the quadratic character $\chi$ mod $p$, which is $\chi(0)=0$, $\chi(j)=+1$ for $j$ a square modulo $p$, and $\chi(j)=-1$ for $j$ a non-square mod $p$.
For odd $n=p_1...p_k$ a product of distinct primes, products of the trivial characters and/or the quadratic characters modulo the various $p_i$ are the $2^k$ real-valued eigenvectors for the Fourier matrix. 
Modulo higher powers $n=p^m$ of a prime, the non-trivial character $\chi(j)$ still just depends on $j$ mod $p$ and whether it's a square or not, or is $0$, and these can be combined multiplicatively as in the previous example.
A: There are several known eigenvector basis for the DFT using
matrices that commute with the DFT matrix:

*

*Alberto Grunbaum (Tridiagonal Commuting matrix), Journal of Mathematical Analysis and Applications, 1982.

*B. Santhanam and T. S. Santhanam (quantum mechanics in finite dimensions)(Symmetric matrix), Signal Processing, Elsevier, 2008.

*Steiglitz and Dickinson (S matrix approach) (Almost tridiagonal), IEEE Transactions on Signal Processing, Feb. 1982.

