I believe the following is true, but I do not know how to prove it.
Given two sets of distinct non-zero complex numbers $z_1, \dots , z_n$ and $w_1, \dots , w_n$, if
$ \begin{bmatrix} z_{1} & \cdots & z_{n} \\ z_{1}^{2} & \cdots & z_{n}^{2} \\ \vdots & \ddots & \vdots \\ z_{1}^n & \cdots & z_{n}^n \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ \vdots \\ 1 \end{bmatrix} = \begin{bmatrix} w_{1} & \cdots & w_{n} \\ w_{1}^{2} & \cdots & w_{n}^{2} \\ \vdots & \ddots & \vdots \\ w_{1}^n & \cdots & w_{n}^n \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ \vdots \\ 1 \end{bmatrix} $
then $w_1, \dots , w_n$ is a permutation of $z_1, \dots , z_n$.
Any ideas?
Background
Again given distinct non-zero complex numbers $z_1, \dots , z_n$, consider the matrix equation:
$ \begin{bmatrix} 1 & z_{1} & z_{1}^2 & \cdots & z_{1}^{n-1} \\ 1 & z_{2} & z_{2}^2 & \cdots & z_{2}^{n-1} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & z_{n} & z_{n}^2 & \cdots & z_{n}^{n-1} \end{bmatrix} \begin{bmatrix} a_0 \\ a_1 \\ \vdots \\ a_{n-1} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ \vdots \\ 0 \end{bmatrix} $
By the fundamental theorem of algebra, this equation can only be satisfied when the $a_1, \dots , a_n$'s are all zero, and so the matrix is non-singular. Since the complex numbers $z_1, \dots , z_n$ are non-zero, the matrix
$ \begin{bmatrix} z_{1} & z_{1}^2 & \cdots & z_{1}^{n} \\ z_{2} & z_{2}^2 & \cdots & z_{2}^{n} \\ \vdots & \vdots & \ddots & \vdots \\ z_{n} & z_{n}^2 & \cdots & z_{n}^{n} \end{bmatrix} $
is also non-singular, and so its columns form a linearly independent basis for $\mathbb{C}^n$. Considered in this light, the claim above compares two bases formed in this way and says, in effect, if the dot-products of each member of the bases with $[ 1,1, \dots ,1]$ are the same, then the two bases are the same.