Let $U \subset \mathbb{R}^n$ be open. Let $F_1,F_2 \in \mathcal{C}^1(U)$ with flow maps $\phi_t^{F_1}, \phi^{F_2}_t$. The two vector fields are topologically equivalent , denoted by $F_1 \simeq F_2 \, $ if there exists a homeomorphism $h: U \rightarrow U$, mapping orbits of the first system onto orbits of the second system, i.e. $$\forall t \in \mathbb{R}, \ \forall x \in U: \quad \phi^{F_1}_t(x)=h^{-1} \circ \phi^{F_2}_{\tau} \circ h(x),$$ with $ \tau: U \times \mathbb{R} \rightarrow \mathbb{R}, \quad \frac{\partial \tau(x,t)}{\partial t} >0 \quad \forall x \in U $. This means the time direction of the orbits is preserved.
We call a set $S \subset U$ invariant if
$$\forall x \in S: \quad \phi_t(x) \in S \quad \forall t$$
My question is the following: Are invariant sets, invariant under topological equivalence? I.e. let $S$ be an invariant set of $F_1$, is $S$ then also an invariant set of $F_2$?
I know this holds for FPs and limit cycles. But is it generally true for invariant sets?