Suppose one has two $n$-tuples of complex numbers $(c_1,\dots,c_n)$ and $(z_1,\dots,z_n)$ such that all $c_i$, $z_i$ are nonzero, and $$ (c_1z_1,\dots,c_nz_n)=(z_{\sigma(1)},\dots,z_{\sigma(n)}) $$ and $$ (c_1^2z_1,\dots,c_n^2z_n)=(z_{\tau(\sigma(1))},\dots,z_{\tau(\sigma(n))}) $$ for some permutations $\sigma\ne\tau\ne1$.
Does this imply that all $c_i$'s are roots of unity?
Here's an example how this can happen: $\sigma=(123456)$, $\tau=(135)(246)$. Then one has \begin{align} z_1&\overset{c_1}{\longmapsto}z_2\overset{c_1}{\longmapsto}z_4\\ z_2&\overset{c_2}{\longmapsto}z_3\overset{c_2}{\longmapsto}z_5\\ z_3&\overset{c_3}{\longmapsto}z_4\overset{c_3}{\longmapsto}z_6\\ z_4&\overset{c_4}{\longmapsto}z_5\overset{c_4}{\longmapsto}z_1\\ z_5&\overset{c_5}{\longmapsto}z_6\overset{c_5}{\longmapsto}z_2\\ z_6&\overset{c_6}{\longmapsto}z_1\overset{c_6}{\longmapsto}z_3 \end{align} and we get the system for $c_i$'s: $$ c_1c_2=c_6,\quad c_2c_3=c_1,\quad c_3c_4=c_2,\quad c_4c_5=c_3,\quad c_5c_6=c_4,\quad c_6c_1=c_5 $$ and one can express $c_3,c_4,c_5,c_6$ as a product of powers of $c_1,c_2$ and then get $c_1^4=c_2^4=1$, so that all $c_i$'s are roots of unity. For example, the tuple of $c_i$'s can be $(i,i,1,i,-i,-1)$.
The question is: Is there a direct way to see that $c_i$'s are roots of unity, without drawing the graphs of permutations and solving corresponding system for the coefficients?
[Maybe if we denote $z=(z_1,\dots,z_n)$, $C=\operatorname{diag}(c_1,\dots,c_n)$, $P$ the permutation matrix for $\sigma$, and $Q$ the permutation matrix for $\tau\sigma$, then $$ C^{-1}Pz=z=C^{-2}Qz,\quad\text{ i.e. }\quad P^{-1}C^{-1}Qz=z $$ Since $P$ and $Q$ are unitary, this resembles the singular value decomposition. Any implications for the eigenvalues of $C$?]
