I don't have a complete solution, but the following may be helpful.

Change variables by $z_i = \sum_j y_j \xi^{ij}$ where $\xi$ is $m$-th primitive root of $1$. Then the first line equations (I am using mod $m$ notation for the indices unless otherwise stated) become $$ 0=\sum_s z_s z_{2t-s} = \sum_{s,j,k} y_jy_k \xi^{sj+(2t-s)k} =\sum_{jk}y_jy_k \xi^{2tk} (m\delta_j^k) = m \sum_k y_k^2 \xi^{2tk} $$ $$=m\sum_{k=0...m/2-1} (y_k^2+y_{m/2+k}^2)\xi^{2tk}. $$ This impies $y_k^2+y_{m/2+k}^2=0$ for all $k$, so $y_k=\pm I y_{m/2+k}$, which are linear equations on $z$.

Similarly, we get linear equations on $z$ from the second a third line in the original post. The problem is now to assure that these are linearly independent for any choices of signs above and any choices in the second and third lines. Good luck!

I don't have a complete solution, but the following may be helpful.

Change variables by $z_i = \sum_j y_j \xi^{ij}$ where $\xi$ is $m$-th primitive root of $1$. Then the first line equations (I am using mod $m$ notation for the indices unless otherwise stated) become $$ 0=\sum_s z_s z_{2t-s} = \sum_{s,j,k} y_jy_k \xi^{sj+(2t-s)k} =\sum_{jk}y_jy_k \xi^{2tk} (m\delta_j^k) = m \sum_k y_k^2 \xi^{2tk} $$ $$=m\sum_{k=0...m/2-1} (y_k^2+y_{m/2+k}^2)\xi^{2tk}. $$ This impies $y_k^2+y_{m/2+k}^2=0$ for all $k$, so $y_k=\pm I y_{m/2+k}$, which are linear equations on $z$.

Similarly, we get linear equations on $z$ from the second and third line in the original post. The problem is now to assure that these are of rank $m$ for any choices of signs above and any choices in the second and third lines. Good luck!

I don't have a complete solution, but the following may be helpful.

Change variables by $z_i = \sum_j y_j \xi^{ij}$ where $\xi$ is $m$-th primitive root of $1$. Then the first line equations (I am using mod $m$ notation for the indices unless otherwise stated) become $$ 0=\sum_s z_s z_{2t-s} = \sum_{s,j,k} y_jy_k \xi^{sj+(2t-s)k} =\sum_{jk}y_jy_k \xi^{2tk} (m\delta_j^k) = m \sum_k y_k^2 \xi^{2tk} $$ $$=m\sum_{k=0...m/2-1} (y_k^2+y_{m/2+1}^2)\xi^{2tk}. $$ This impies $y_k^2+y_{m/2+k}^2=0$ for all $k$, so $y_k=\pm I y_{m/2+k}$, which are linear equations on $z$.

Similarly, we get linear equations on $z$ from the second a third line in the original post. The problem is now to assure that these are linearly independent for any choices of signs above and any choices in the second and third lines. Good luck!

I don't have a complete solution, but the following may be helpful.

Change variables by $z_i = \sum_j y_j \xi^{ij}$ where $\xi$ is $m$-th primitive root of $1$. Then the first line equations (I am using mod $m$ notation for the indices unless otherwise stated) become $$ 0=\sum_s z_s z_{2t-s} = \sum_{s,j,k} y_jy_k \xi^{sj+(2t-s)k} =\sum_{jk}y_jy_k \xi^{2tk} (m\delta_j^k) = m \sum_k y_k^2 \xi^{2tk} $$ $$=m\sum_{k=0...m/2-1} (y_k^2+y_{m/2+k}^2)\xi^{2tk}. $$ This impies $y_k^2+y_{m/2+k}^2=0$ for all $k$, so $y_k=\pm I y_{m/2+k}$, which are linear equations on $z$.

Similarly, we get linear equations on $z$ from the second a third line in the original post. The problem is now to assure that these are linearly independent for any choices of signs above and any choices in the second and third lines. Good luck!