Consider a reaction-diffusion system \begin{equation}\tag{1}\label{(1)} u_t=Du_{xx}+F(u)\in\mathbb{R}^N. \end{equation} Making the modulated travelling wave ansatz \begin{equation} u=u(\xi,\tau)\text{ with }\xi=x-ct\text{ and }\tau=\omega t, \end{equation} I get the so-called modulated travelling wave equation \begin{equation} u_{\xi\xi}=D^{-1}(-cu_{\xi}+\omega u_{\tau}-F(u)) \end{equation} which as a first-order system in the spatial variable $\xi$ can be written as \begin{align}\tag{2}\label{(2)} &u_{\xi}=v\\ &v_{\xi}=D^{-1}(-cv+\omega u_{\tau}-F(u)). \end{align}
Regard system $\eqref{(2)}$ as a dynamical system with an underlying rotational symmetry acting through the temporal shift \begin{equation} v(\xi,\tau)\mapsto \Gamma(\varphi)(v)=v(\xi,\tau+\varphi), \varphi\in S^1. \end{equation}
Next, two statements are made which I think I do not fully understand.
(a) Spatially homogeneous equilibria of equation $\eqref{(1)}$ correspond to equilibria of the system $\eqref{(2)}$ contained in the fixed point space $\text{Fix}(S^1)$.
(b) Spatially periodic, temporally steady solutions of equation $\eqref{(1)}$ correspond to periodic orbits, or, more precisely, relative equilibria of system $\eqref{(2)}$ with respect to the $S^1$-symmetry: $$ v(\xi,\tau)=v_{*}(x)=v_{*}\left(\xi+\frac{c}{\omega}\tau\right). $$
My explanations:
I am not sure if they make sense! Anyway, I post them here in order to get some feedback or help.
(a):
For simplicity, let $D=1$. Let $\tilde{u}=\tilde{u}(x,t)$ be a spatially homogeneous equilibrium of equation $\eqref{(1)}$: $$ \tilde{u}_t=D\underbrace{\tilde{u}_{xx}}_{=0}+F(\tilde{u})=0\implies F(\tilde{u})=0. $$ For the system $\eqref{(2)}$, \begin{align} &u_{\xi}=v\\ &v_{\xi}=-cv+\omega u_{\tau}-F(u), \end{align} the solution $(\tilde{u},\tilde{u}_{\xi})(\xi,\tau)$ is an equilibrium if $v=\tilde{u}_{\xi}=0$ (since then the first equation is zero) and $\tilde{u}_{\tau}=0$ (since then the second equation is zero).
But, as far as I see, $$ \tilde{u}_{\xi}=0=\tilde{u}_{\tau}\implies \tilde{u}(\xi,\tau)\in\text{Fix}(S^1) $$ since otherwise (i.e. if $\Gamma(\tilde{u}(\xi,\tau))=\tilde{u}(\xi,\tau+\varphi)$ for some $\varphi\in S^1$), $\tilde{u}_{\tau}\neq 0$.
(b):
Again, let $D=1$ for simplicity. Let $\tilde{u}=\tilde{u}(x,t)$ be a spatially periodic, temporally steady solution of equation $\eqref{(1)}$.
Then $$ u(\xi,\tau)=(\tilde{u},v=\tilde{u}_{\xi})(\xi,\tau)=\left(\tilde{u}\left(\xi+\frac{c}{\omega}\tau\right),v\left(\xi+\frac{c}{\omega}\tau\right)\right)=(\tilde{u}(x),v(x)) $$ is a solution of the system $\eqref{(2)}$ if the spatial periodicity of $\tilde{u}$ is $\tilde{u}(x)=\tilde{u}(x+L)$ with $L=\frac{2\pi}{k}, k=\frac{\omega}{c}$.
In particular, $$ \tilde{u}(x)=\tilde{u}\left(\xi+\frac{c}{\omega}\tau\right) $$ is a travelling wave solution in the coordinates $(\xi,\tau)$ and hence a wave train since $\tilde{u}(x)$ is spatially periodic by assumption. Hence, in the phase space, $\tilde{u}(x)$ is a periodic orbit.
Moreover, as a travelling wave in the coordinates $(x=\xi+\frac{c}{\omega}\tau,\tau)$, the solution $\tilde{u}(x)$ does not depend on the second coordinate $\tau$ and hence the translation map $\Gamma$ has no influence on the solution $\tilde{u}(x)$.
In other words: With respect to the $S^1$-symmetry, acting through $\Gamma$ on the second coordinate $\tau$, $\tilde{u}$ is an equilibrium since nothing happens with $\tilde{u}$ if the symmetry is acting.