This was asked in Math Stackexchange here but generated no comments or answers. I have slightly edited the original question with the comment in the fourth paragraph and the explicit matrix example at the end.

I have a very concrete question. First an example: If one considers the additive group $Z/(n)$ which is cyclic, the corresponding group characters are the rows of the discrete fourier transform (DFT)matrix $D$.

Take any $n\times n$ circulant matrix $C$. Obviously this matrix is generated by the action of the cyclic group on the first row. Its eigenvectors are the columns of $D$ (which is symmetric) and the matrix of eigenvectors diagonalizes any circulant matrix. Corresponding to each root of unity $\omega$ of order $n,$ its consecutive powers make up the eigenvector and the inner product of the first row of $C$ is clearly the corresponding eigenvalue.

Now comes the question. For simplicity, assume that $n$ is a prime. Consider the multiplicative group of $Z/(n)$ which has $n-1$ elements. The Dirichlet characters of this group form an $(n-1)\times(n-1)$ matrix $D'$. Now take a ``multiplicative matrix'', $M$ where first one specifies the first row $(a_1,\ldots,a_{n-1})$ and then obtains the $k^{th}$ row as $(a_{k},a_{2k},\ldots,a_{(n-1)k})$ where the indices are computed modulo $n.

In fact, if $n$ is assumed to be prime, I'd expect that the eigenvectors to be obtained by a coordinate permutation of the eigenvectors which are the columns of the DFT matrix since both groups in question are cyclic.

I'd have assumed that such a matrix would have exactly the Dirichlet characters, of which there are enough when $n$ is a prime, as its eigenvectors. A few basic computations on Mathematica seemed to contradict this hope. For example taking $n=7,$ and the first row as $(1,1,1,0,0,0)$ yielding the matrix $$ \left( \begin{array}{cccccc} 1 &1 &1 & 0& 0& 0\\ 0 &1 &0 & 1& 0& 1\\ 0 &1 &1 & 0& 0& 1\\ 1 &0 &0 & 1& 1& 0\\ 1 &0 &1 & 0& 1& 0\\ 0 &0 &0 & 1& 1& 1 \end{array} \right) $$ the expected eigenvectors materialized. However, taking the first row as $(1,1,0,0,0,0)$ yielding the matrix $$ \left( \begin{array}{cccccc} 1 &1 &0 & 0& 0& 0\\ 0 &1 &0 & 1& 0& 0\\ 0 &0 &1 & 0& 0& 1\\ 1 &0 &0 & 1& 0& 0\\ 0 &0 &1 & 0& 1& 0\\ 0 &0 &0 & 0& 1& 1 \end{array} \right) $$ did not.

Why is this the case? References, pointers welcome.

This was asked in Math Stackexchange here but generated no comments or answers. I have slightly edited the original question with the comment in the fourth paragraph and the explicit matrix example at the end.

I have a very concrete question. First an example: If one considers the additive group $Z/(n)$ which is cyclic, the corresponding group characters are the rows of the discrete fourier transform (DFT)matrix $D$.

Take any $n\times n$ circulant matrix $C$. Obviously this matrix is generated by the action of the cyclic group on the first row. Its eigenvectors are the columns of $D$ (which is symmetric) and the matrix of eigenvectors diagonalizes any circulant matrix. Corresponding to each root of unity $\omega$ of order $n,$ its consecutive powers make up the eigenvector and the inner product of the first row of $C$ is clearly the corresponding eigenvalue.

Now comes the question. For simplicity, assume that $n$ is a prime. Consider the multiplicative group of $Z/(n)$ which has $n-1$ elements. The Dirichlet characters of this group form an $(n-1)\times(n-1)$ matrix $D'$. Now take a ``multiplicative matrix'', $M$ where first one specifies the first row $(a_1,\ldots,a_{n-1})$ and then obtains the $k^{th}$ row as $(a_{k},a_{2k},\ldots,a_{(n-1)k})$ where the indices are computed modulo $n.

In fact, if $n$ is assumed to be prime, I'd expect that the eigenvectors to be obtained by a coordinate permutation of the eigenvectors which are the columns of the DFT matrix since both groups in question are cyclic.

I'd have assumed that such a matrix would have exactly the Dirichlet characters, of which there are enough when $n$ is a prime, as its eigenvectors. A few basic computations on Mathematica seemed to contradict this hope. For example taking $n=7,$ and the first row as $(1,1,1,0,0,0)$ yielding the matrix $$ \left( \begin{array}{cccccc} 1 &1 &1 & 0& 0& 0\\ 0 &1 &0 & 1& 0& 1\\ 0 &1 &1 & 0& 0& 1\\ 1 &0 &0 & 1& 1& 0\\ 1 &0 &1 & 0& 1& 0\\ 0 &0 &0 & 1& 1& 1 \end{array} \right) $$ the expected eigenvectors materialized. However, taking the first row as $(1,1,0,0,0,0)$ yielding the matrix $$ \left( \begin{array}{cccccc} 1 &1 &0 & 0& 0& 0\\ 0 &1 &0 & 1& 0& 0\\ 0 &0 &1 & 0& 0& 1\\ 1 &0 &0 & 1& 0& 0\\ 0 &0 &1 & 0& 1& 0\\ 0 &0 &0 & 0& 1& 1 \end{array} \right) $$ did not.

Why is this the case? References, pointers welcome.