This is a partial answer. What you are looking for is an eigenvector $h$ of the matrix $M$ whose entries are $\omega^{(j-1)(k-1)}$. The eigenvalue is $c$. Because $M^*M=nI_n$, we must have $|c|=\sqrt n$. because of the first equation $R(1)=ch_1$ and the fact that $h_j=\pm1$, we see that $c$ must be real. Hence $c=\pm\sqrt n$. Now the question reduces to whether $Mh=\pm nh$ has a solution with $h_j=\pm1$. One way to continue the analysis is to calculate the powers $M^2,\ldots$. They look to be much simpler than $M$ itself, due to cancellations, and we have $M^kh=c^kh$. For instance $M^2h=nh$ and $M^4h=n^2h$. Whence necessary conditions.

More precisely, $N=M^2$ has entries $n_{jk}=0$, except for $j+k=2$ (mod $n$). Its eigenvalues are $\pm n$, with respective multiplicities $2k+1$ and $2k-1$. Only $c^2=n$ matters for you. If $c=\sqrt n$ then you have $h_j=h_{n+2-j}$ for every $j=2,\ldots,2k$. If $c=-\sqrt n$, you have instead $h_j=-h_{n+2-j}$.

This is a partial answer. What you are looking for is an eigenvector $h$ of the matrix $M$ whose entries are $\omega^{(j-1)(k-1)}$. The eigenvalue is $c$. Because $M^*M=nI_n$, we must have $|c|=\sqrt n$. because of the first equation $R(1)=ch_1$ and the fact that $h_j=\pm1$, we see that $c$ must be real. Hence $c=\pm\sqrt n$. Now the question reduces to whether $Mh=\pm nh$ has a solution with $h_j=\pm1$. One way to continue the analysis is to calculate the powers $M^2,\ldots$. They look to be much simpler than $M$ itself, due to cancellations, and we have $M^kh=c^kh$. For instance $M^2h=nh$ and $M^4h=n^2h$. Whence necessary conditions.

More precisely, $N=M^2$ has entries $n_{jk}=0$, except for $j+k=2$ (mod $n$). Its eigenvalues are $\pm n$, with respective multiplicities $2k+1$ and $2k-1$. Only $c^2=n$ matters for you. Then you have $h_j=h_{n+2-j}$ for every $j=2,\ldots,2k$.

This is a partial answer. What you are looking for is an eigenvector $h$ of the matrix $M$ whose entries are $\omega^{(j-1)(k-1)}$. The eigenvalue is $c$. Because $M^*M=nI_n$, we must have $|c|=\sqrt n$. because of the first equation $R(1)=ch_1$ and the fact that $h_j=\pm1$, we see that $c$ must be real. Hence $c=\pm\sqrt n$. Now the question reduces to whether $Mh=\pm nh$ has a solution with $h_j=\pm1$. One way to continue the analysis is to calculate the powers $M^2,\ldots$. They look to be much simpler than $M$ itself, due to cancellations, and we have $M^kh=c^kh$. For instance $M^2h=nh$ and $M^4h=n^2h$. Whence necessary conditions.

This is a partial answer. What you are looking for is an eigenvector $h$ of the matrix $M$ whose entries are $\omega^{(j-1)(k-1)}$. The eigenvalue is $c$. Because $M^*M=nI_n$, we must have $|c|=\sqrt n$. because of the first equation $R(1)=ch_1$ and the fact that $h_j=\pm1$, we see that $c$ must be real. Hence $c=\pm\sqrt n$. Now the question reduces to whether $Mh=\pm nh$ has a solution with $h_j=\pm1$. One way to continue the analysis is to calculate the powers $M^2,\ldots$. They look to be much simpler than $M$ itself, due to cancellations, and we have $M^kh=c^kh$. For instance $M^2h=nh$ and $M^4h=n^2h$. Whence necessary conditions.

More precisely, $N=M^2$ has entries $n_{jk}=0$, except for $j+k=2$ (mod $n$). Its eigenvalues are $\pm n$, with respective multiplicities $2k+1$ and $2k-1$. Only $c^2=n$ matters for you. If $c=\sqrt n$ then you have $h_j=h_{n+2-j}$ for every $j=2,\ldots,2k$. If $c=-\sqrt n$, you have instead $h_j=-h_{n+2-j}$.