Do equivariant dynamics have invariant equilibria? Let $\phi_t$ be a one-parameter subgroup of diffeomorphisms of a manifold ($\mathbb{R}^n$ for simplicity). In other words, $\varphi$ is a continuous dynamics.
Suppose that $\varphi_t$ is $G$-equivariant for every $t$, where $G$ is a Lie group. Is is true that the equilibria of $\varphi$ (i.e., the points fixed by $\varphi_t$ for every $t$) are $G$-invariant i.e., fixed by all the elements of $G$? Is is true at least for stable equilibria of $\phi$?
I think this might be related to 'spontaneous symmetry breaking'.
Thank you in advance.
 A: The short answer is "it depends". If the symmetry group $G$ is discrete, then the answer is "no". Take a system $\dot{x} = x ( a - x^2 - b y ^2), \; \dot{y} = y ( a - b x^2 - y^2 )$. Its symmetry group $G$ is generated by reflections $(x,y) \mapsto (-x, y)$ and $(x, y) \mapsto (x, -y)$. You can choose coefficients $a$ and $b$ in such way that there would be two pairs of hyperbolic sinks with lesser isotropy group (i.e. they are not invariant under the action of all elements of $G$). This system is basically a restriction of Guckenheimer-Holmes example to one of the coordinate planes.
If we don't assume hyperbolicity of equilibria in case of continuous $G$, there is another simple counter-example. Take a planar system which in polar coordinates can be written as $\dot{r} = 0, \; \dot{\varphi} = f(r)$, where $f(0) = f(1) = 0$. The phase space is foliated by invariant circles, on most of which the dynamics is just a rotation. However, there is an equilibrium at the origin and there is a circle of non-hyperbolic equilibria at $r = 1$. The symmetry group here is the rotation group $SO(2)$. Clearly, while equilibrium at the origin is invariant under the action of group $G$, any individual equilibrium at circle $r = 1$ is not: any of these can be obtained from another one by action of some element from group $G$.
However, if we assume that system in consideration has only hyperbolic equilibria, the answer to your question seems to be "yes". The structure of a hyperbolic equilibrium is well-known due to Grobman-Hartman theorem. Namely, hyperbolic equilibria are always isolated: there is a neighbourhood $U_{x^\ast}$ of equilibrium $x^\ast$ that contains no other equilibria. If $u(t) \equiv x^\ast$ is an equilibrium solution, $\gamma \cdot u(t) = \gamma \cdot x^\ast$ is also a solution and it is an equilibrium one too. However, due to continuity of group action $G$ there should be some group element $\hat{\gamma}$ close to identity such that $\hat{\gamma} \cdot x^\ast \in U_{x^\ast}$ and is also an equilibrium. That contradicts the fact that $x^\ast$ is an isolated equilibrium.
As a side note: as far as I understand, there are methods to deal with non-hyperbolic equilibria in equivariant systems if the only reason for their non-hyperbolicity is a symmetry group action. It is called symmetry reduction and described in books

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*M. Golubitsky, I. Stewart, and D. G. Schaeffer (1988) Singularities and Groups in Bifurcation Theory, Volume II. Springer-Verlag, New York.

*P. Chossat and R. Lauterbach (2000) Methods in Equivariant Bifurcations and Dynamical Systems. World Scientific, Singapore.

