Consider the sets $\mathfrak{X}(\mathbb{R}^3) = \{X: \mathbb{R}^3 \to \mathbb{R}^3; X \mbox{ is smooth}\}$ and $\Sigma = \{0\}\times\mathbb{R}^2$.
Let $X, Y$ be vector fields in $\mathfrak{X}(\mathbb{\mathbb{R}}^3)$, such that
$$X(x,y,z) = (y,z,1) $$ and $$Y(x,y,z) = (y+ \mathcal{O}_1(2),z+\mathcal{O}_2(2),1), $$ where $\mathcal{O}_i(3):\mathbb{R}^3 \to \mathbb{R}$ is a smooth function satisfying
$$\exists \ r_i>0, \exists \ K_i>0; \ |\mathcal{O}_i(2)(x,y,z)|\leq K_i\|(x,y,z)\|^2,\ \forall\ (x,y,z) \in B_{r_i}(0,0,0). $$
Definition: Let $Z_1,Z_2$ $\in$ $\mathfrak{X}{(\mathbb{R}^3)}$, we say that $Z_1,Z_2$ are $\Sigma$-equivalent at the origin if there exists a homeomorphism $h:U_0\to V_0$, where $U_0$ and $V_0$ are neighborhoods of the origin such that
$h(U_0\cap \Sigma) = V_0 \cap \Sigma$
$h$ maps the orbits of $Z_1$ into orbits of $Z_2$ preserving the orientation of the orbits.
My question:
Question: Are $X, Y$ (as defined at the beginning of the text) $\Sigma$-equivalent at the origin?
Just for the records, these orbits are in respect of the ODEs $$(\dot{x},\dot{y},\dot{z})= X(x,y,z) $$ and $$(\dot{x},\dot{y},\dot{z})= Y(x,y,z). $$
Moreover, we can assume that $$Y(x,0,z) = (0,z+ \mathcal{O(2)}, 1)\ \text{and} \ Y(x,y,0) = (y+\mathcal{O}(2),0,1).$$
It seems true, however, I don't have many ideas of how construct such $h$, can anyone help me?