Let $M$ be a compact manifold and we are given two VFs $F,F' \in \mathcal{X}(M)$. We define a DS by $$\dot{x}=F(x)$$ and equivalently for $F'$.
The two vector fields are topologically equivalent , denoted by $F \simeq F' \, $ if there exists a homeomorphism $h: U \rightarrow U$, mapping orbits of the first system onto orbits of the second system, i.e. $$\forall t \in \mathbb{R}, \ \forall x \in U: \quad \phi^{F}_t(x)=h^{-1} \circ \phi^{F'}_{\tau} \circ h(x),$$ 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 make $\mathcal{X}(M)$ a topological space by using the $\mathcal{C}^1$-induced topology.
My question is the following: Can we define a function $$g: \mathcal{X}(M) \times \mathcal{X}(M) \rightarrow \mathbb{R}$$ which is continuous in the $\mathcal{C}^1$ induced topology and 0 if and only if the two VFs are topologically equivalent, i.e. $$ g(F,F')=0 \quad \Leftrightarrow \quad F \simeq F'$$
No such function $g$ exists in general, and in fact the answer I gave to your previous question can be used to produce a counterexample.
In that answer I gave an example of a family of flows $\phi_r$ on the torus $T^2 = S^1 \times S^1$, defined for all $r \in \mathbb R \cup \infty$. If we think of $\mathbb R \cup \infty \approx S^1$, and if we define the vector field $F_r$ so that the flow $\phi_r$ that it generates is a unit speed flow on $T^2$ (i.e. if we define $F_r$ so that all its vectors have length $1$), then this family $F_r$ is continuous in the $C^1$ induced topology. And as mentioned in that answer, $F_r$ and $F_s$ are topologically equivalent if and only if the numbers $r,s$ are in the same orbit as the action of the group $\text{SL}_2(\mathbb Z)$ --- which, by definition, means that there exists a matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ with integer coefficients such that $ad-bc=1$ and such that $r=\frac{as+b}{cs+d}$.
However, each orbit of the action of $\text{SL}_2(\mathbb Z)$ on $\mathbb R \cup \{\infty\}$ is dense in $\mathbb R \cup \{\infty\}$. So for any ordered pair $(t,u) \in (\mathbb R \cup \{\infty\})^2$, there exists a sequence of ordered pairs $(r_n,s_n)$ such that the entire set $\bigcup_{n \in \mathbb Z} \{r_n,s_n\}$ is in a single orbit of the $\text{SL}_2(\mathbb Z)$ action, and such that $\lim r_n = t$ and $\lim s_n = u$. So if $g$ is continuous in the $C^1$ induced topology and is zero whenever $F_r,F_s$ are topologically equivalent, it follows that $g(r_n,s_n)=0$. By continuity of $g$ it follows that $g(t,u)=0$. So $g$ cannot distinguish when $F_t$, $F_u$ are topologically inequivalent.