Making the matrix $A_{mn} = \frac{1}{\sqrt{N}} e^{-i\frac{2\pi}{N}x_m k_n}$ unitary Let's say we have a matrix
$$A_{mn} = \frac{1}{\sqrt{N}}e^{-i\frac{2\pi}{N} x_m k_n}$$
where $0 \leq x_m < N$, and each of $m, n$ is an integer from $0$ to $N - 1$. If $x_m$ is an integer where $x_m = m$, we can find that $k_n = n$ makes the matrix $A_{mn}$ unitary, i.e., $\mathbf{AA}^H = \mathbf{A}^H\mathbf{A} = \mathbf{I}$.
However, if $x_m$ is not integer (but assuming the values of $x_m$ are all unique), does there exist a set of $k_n$ that could make the matrix $A_{mn}$ unitary? If it exists, how to find $k_n$?

A suitable set of $k_n$ does not necessarily exist; here's a counterexample for $N=3$. Take
$$
x_0 = 0, \quad x_1 = \frac 13, \quad x_2 = 1.
$$
The orthogonality of the first two columns gives us
$$
0 = \sum_{n=0}^2 e^{-i\frac{2\pi}{3} k_n(x_1 - x_0)} 
=  \sum_{n=0}^2 e^{-i\frac{2\pi}{3} \frac 13 k_n} = \sum_{n=0}^2 e^{-i\frac{2\pi}9 k_n}
$$
Adding in the result from the other two pairs gives us
$$
0 = \sum_{n=0}^2 e^{-i\frac{2\pi}9 k_n}= 
\sum_{n=0}^2 e^{-i\frac{4\pi}9 k_n}=
\sum_{n=0}^2 e^{-i\frac{6\pi}9 k_n}.
$$
Take $z_n = e^{-i\frac{2\pi}9 k_n}$; note that $|z_n| = 1$. The above equations can be rewritten as
$$
0 = z_0 + z_1 + z_2 = z_0^2 + z_1^2 + z_2 ^2 = z_0^3 + z_1^3 + z_2^3.
$$
With Newton's identities, we can deduce that the unique solution to this system of equations is $z_0 = z_1 = z_2  = 0$. Our $z_n$ are non-zero, so there are no suitable $k_n$.
