Let $M$ be a smooth manifold, and $\mathfrak{X}$ be a global smooth vector field. Let $D \subseteq \mathbb{R} \times M$ be the flow domain of the maximal flow $\Phi$ of $\mathfrak{X}$, and for any $p \in M$ let $D_p = \{ t \in \mathbb{R}: (t,p) \in D \}$ and $\mathcal{O}_p = \Phi(D_p)$ (i.e., the maximal orbit of $p$).
It is easy to come up with examples showing that, in general, if $N$ is a smooth embedded submanifold of $M$ then the set $$ \mathcal{V} = \bigcup_{p \in N} \mathcal{O}_p $$ is not a smooth embedded, or even immersed, submanifold. However, the only examples I've come up with involve $N$'s on which $\mathfrak{X}$ fails to be tangent.
I was wondering to what extent, if at all, the situation is improved by having $\mathfrak{X}$ everywhere tangent to $N$. Is this sufficient to demonstrate that $\mathcal{V}$ is an immersed submanifold of $M$? Or even an embedded one? Are there simple further conditions one can impose on the pair $(N, \mathfrak{X})$ to ensure that $\mathcal{V}$ is even embedded? If $\Phi(D_p) \subseteq N$ for all $p \in N$ then this is trivial, so I'm concerned about the situation where $N$ is not invariant under the maximal flow.
Any help or pointer to a reference would be much appreciated!
Consider the torus $T^2$, it is the quotient of $R^2$ by the group $G$ generated by two translations $t_a, t_b$ of respective directions $a$ and $b$, consider $\alpha$ an irrational number. The vector field $X(x)=a+\alpha b$ on $R^2$ is invariant by $G$, it induces a vector $Y$ on $X$. The orbits of $Y$ are dense in $T^2$. Take $x\in T^2$, you have a morphism $f:[0,1]\rightarrow T^2$ such that $f$ is injective and $f([0,1])$ is contained in the orbit of $x$ by the flow of $Y$, the saturated space $V$ of $N=f((0,1))$ is the orbit of $x$ so it is a dense subset of $T^2$, thus it is not a submanifold of $T^2$.