characterizing equations above reduce to $$\begin{aligned}
ce^{i{\theta}(2{\theta}-1)t}  & = 1  \\
ce^{i{\theta}t} & = -1 \\
ce^{-i{\theta}t}  & = 1.\end{aligned}$$ Thus $c=e^{i{\theta}t} = \pm i$, so ${\theta}t = \pi(m+1/2)$ for some integer $m$, and $t = \pi(m+1/2)/{\theta}$ and $kt = (2{\theta}-1)\pi(m+1/2)$. When ${\theta}$ is even, any $m$ is a valid solution and uniform mixing occurs. These are the Latin square graphs $L_{\theta}(2{\theta})$. When ${\theta}$ is odd, no solutions occur.

For the complement, the characterizing equations for $\overline{G}$ reduce to $$\begin{aligned}
ce^{i(2{\theta}^2+{\theta}-1)t} & = 1  \\
ce^{i({\theta}-1) t} & = 1  \\
ce^{i(-{\theta}-1)t} & = -1\end{aligned}$$ for some ${\left|c\right|} = 1$. Thus $c = e^{-i({\theta}-1) t}$ and the first and third equations are $e^{it2{\theta}^2t} = 1$ and $e^{-i2{\theta}t} = -1$. Thus $e^{i{\theta}t} = \pm i$, so ${\theta}t = \pi(m+1/2)$ for some $m$, and $t = \pi(m+1/2)/{\theta}$. When ${\theta}$ is even, any $m$ is a valid solution and uniform mixing occurs. When ${\theta}$ is odd, no solutions occur.

If $X$ or $\overline{X}$ has parameters $(4{\theta}^2,2{\theta}^2+ {\theta}, {\theta}^2+{\theta},{\theta}^2+{\theta})$, then a similar results holds: there are no solutions for ${\theta}$ even, and for ${\theta}$ odd $t = \pi(m+1/2)/{\theta}$ is a solution for every $m$. These are the negative Latin square graphs $NL_{\theta}(2{\theta})$. The same 