Equivariant implicit function theorem Let $f:\mathbb{R}\times \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$ be a smooth function and $G\subset \operatorname{SO}(n)$ be a $1$-dimensional compact Lie group (diffeomorphic to the circle). Moreover let $G$ act on $\mathbb{R}^{n}$ by standard left multiplication. We assume that $f$ is equivariant with respect to $G$, i.e. for all $g\in G$ and all $(t,x)\in \mathbb{R}\times \mathbb{R}^{n}$ we have $f(t,g\cdot x) = g\cdot f(t,x)$. Let now $(t_{0},x_{0}) \in \mathbb{R}\times \mathbb{R}^{n}$ such that $x_{0} \not= 0$, $f(t_{0},x_{0}) = 0$ and
$$\ker \left ( \frac{\partial f}{\partial x}(t_{0},x_{0}) \right ) = T_{x_{0}}(G\cdot x_{0}),$$
i.e. the kernel of the Jacobian of $f$ is only in the direction of the action.
QUESTION: Is there any version of the implicit function theorem in this setting? Does one need more additional conditions to be able to construct near-by solutions? If yes, which conditions are these?
Thanks in advance.
 A: The equivariant version of the implicit function theorem is the following.

Let $f: \mathbb{R}^p \times \mathbb{R}^n \to \mathbb{R}^m$ be a smooth function (possibly only defined on open neighborhoods) which is equivariant with respect to the action of a compact Lie group $G$ on $\mathbb{R}^n$ and $\mathbb{R}^m$. Let $(t_0, x_0)$ be such that $f(t_0, x_0) = 0$. Assume that the stabilizer of $x_0$ is trivial and that $0$ is a fix point for the action on $\mathbb{R}^m$ (these assumptions are not essential but simplify the argument, see below). Consider the so-called deformation complex $$\mathfrak{g} \to \mathbb{R}^n \to \mathbb{R}^m,$$
where the first arrow is the action of the Lie algebra at the point $x_0$, i.e. $\xi \mapsto \xi \,. x_0$ and the second arrow is the differential of $f$ at $(t_0, x_0)$ with respect to the second slot (i.e. the Jacobian).
If this complex is exact, then there exist a smooth function $x: \mathbb{R}^p \to \mathbb R^n$ such that
$$
\{ (t, g \cdot x(t)) | t \in \mathbb R^p, g \in G \} = f^{-1}(0).
$$

(I'm a bit sloppy here and in the proof below: everything needs to be restricted to open neighborhoods of $t_0$, $e \in G$ and $0 \in \mathbb{R}^m$ etc).
Remark: Your assumption about the derivative is equivalent to the exactness of the complex in the first arrow. However, since you assume that $n = m$, the complex is never exact in the second arrow. What you could do is apply this result to the function $pr \circ f$, where $pr$ is the projection onto the image of the Jacobian.
Proof: Since $G$ is compact, and the action is free at $x_0$, there exist slice coordinates around $x_0$, i.e. there is a map $\iota: \mathbb{R}^{n-d} \to \mathbb{R}^n$ such that $\iota(0) = x_0$ and such that $\chi: G \times \mathbb{R}^{n-d} \to \mathbb{R}^n, (g, y) \mapsto g \cdot \iota(y)$ is a local diffeo (here $d$ is the dimension of $G$). Define the map $F: \mathbb{R}^p \times \mathbb{R}^{n-d} \to \mathbb{R}^{n}$ by $F(t, y) = f(t, \iota(y))$. The assumption about the exactness of the deformation complex is equivalent to the invertibility of the Jacobian of $F$. Thus, using the ordinary implicit function theorem, there exist $y(t)$ such that $F(t, y(t)) = 0$ and every such point in the zero level set is of this form. Set $x(t) = \iota(y(t))$ and the claim follows as $\chi$ is a local diffeo.
Final remark: The statement generalizes directly to actions on manifolds, and properness of the action is enough (instead of compactness of $G$). In fact, they generalize even to the infinite-dimensional setting. Moreover, the assumptions about the stabilizers of $x_0$ and $f(t_0, x_0)$ can be relaxed. You can find the details in my PhD thesis https://arxiv.org/abs/1909.00744 and in the paper https://arxiv.org/abs/2010.10165.
