My goal is to prove that any O(2)-equivariant system of ordinary differential equations in the plane can be written in the form
$$\begin{pmatrix} \dot{x} \\ \dot{y} \end{pmatrix} =\left[ \lambda +g(x^2+y^2) \right]\begin{pmatrix} x \\ y \end{pmatrix} $$
In order to be O(2)-equivariant, a function $\zeta:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ must satisfy the relation
$$\zeta(\psi(\epsilon,(x,y)))=\psi(\epsilon,\zeta(x,y)),$$ where $\psi(\epsilon,(x,y))$ is the group action, for all $\epsilon$ (the group parameter) and any $(x,y)\in\mathbb{R}^2$. My thought is that I can construct the general function $\zeta(x,y)=(\zeta_1(x,y),\zeta_2(x,y))^T=\left[ \lambda +g(x^2+y^2) \right]\begin{pmatrix}x, & y\end{pmatrix}^T$ by tackling reflection symmetry and continuous rotation symmetry separately.
Looking at reflection first, it's obvious that if a function $\zeta(x,y)=(\zeta_1(x,y),\zeta_2(x,y))^T$ is to be equivariant under reflection about the $y$-axis, then $\zeta_1(-x,y)=-\zeta_1(x,y)$ and $\zeta_2(-x,y)=\zeta_2(x,y)$. That is, $\zeta_1(x,y)$ is odd in $x$ while $\zeta_2(x,y)$ is even in $x$.
Now considering the continuous rotation about the origin, I think (but have not proven) that it would be sufficient to show
$$\frac{d}{d\epsilon}\bigg\rvert_{\epsilon=0}\zeta(\psi(\epsilon,(x,y)))=\frac{d}{d\epsilon}\bigg\rvert_{\epsilon=0}\psi(\epsilon,\zeta(x,y))$$ $$\Downarrow$$ $$\boldsymbol{v}(\zeta)(x)=\boldsymbol{v}\bigg\rvert_{(x,y)=\zeta(x,y)}$$
where $\boldsymbol{v}=y\partial_x-x\partial_y$ is the infinitesimal generator of the O(2). I found that this equation determines a first-order system of PDEs,
$$\zeta_2=y\partial_x\zeta_1-x\partial_y\zeta_1$$ $$-\zeta_1=y\partial_x\zeta_2-x\partial_y\zeta_2$$
which does not seem right. Am I completely off base here? I'm new to Lie groups and algebras so I'm open to all kinds of criticism.