Let $F:U\rightarrow \Bbb{R}^d,G:V\rightarrow \Bbb{R}^d$ smooth vector fields, where $U,V$ are open sets in $\Bbb{R}^d$. Let $h:U\rightarrow V$ a diffeomorphism such that
$$dh(x)F(x)=G(h(x))\quad\forall x\in U $$
I want to prove that
$$h(\omega_F(x))=\omega_G(h(x)) \quad\forall x\in U, $$
where $\omega_F$ and $\omega_F$ are the $\omega$-limit sets for $F$ and $G$ respectively.
I managed to prove that, if $\varphi(t,x)$ is an orbit of $F$, so $\psi(t,x):=h(\varphi(t,x))$ is an orbit of $G$, and I'm trying to use it to prove that exists a sequence $t_n\nearrow \infty$ such that
$$\lim \varphi(t_n,x)=h(p), $$
if, and only if, there is a sequence $t_n\nearrow \infty$ such that
$$\lim \psi(t_n,h(x))=p. $$
How can I do it?
Note that $$\frac{d}{dt}\bigg|_{t=0} \psi(t,x) = dh(\varphi(0,x)) \frac{d}{dt}\bigg|_{t=0} \varphi(t,x) = dh(x) F(x) = G(h(x)) \ne G(x).$$ Thus, $\psi$ actually is not the correct flow of the vector field $G$. What you need instead is $$\psi(t,x) = h(\varphi(t,h^{-1}(x))).$$ With this corrected version of $\psi$, you should be able to do it by plugging in the formula.
The technical term is that $h$ is a conjugacy between the dynamical systems $\varphi$ and $\psi$ (and the two systems are called conjugate). Morally, two conjugate systems are the same from a dynamical point of view.