Let $M$ be a compact $n$-manifold (let's say with boundary, but this isn't too important), $X$ a vector field on $M$ and $f:M \to \mathbb{R}$ a non-zero function on $M$.
My question: are the singular foliations spanned by $X$ and $fX$ equal? In other words, do the traces of trajectories of $X$ and $fX$ with the same starting point coincide?
I tried a couple of simple examples and checked that they do coincide, and I did a heuristic argument which works in my favor: if $p \in M$ such that $X_p = 0$, then obviously the trajectories starting at $p$ of $X$ and $fX$ coincide (it's just the point $p$). Otherwise, there exists a chart centered at $p$ where $X = \frac{\partial}{\partial x_1}$, and so in this chart, $\phi_t^X(p) = (e^t,0,\dots,0)$. On the other hand, the flow of $fX$ starting at $p$ is locally the solution of the equation $$x' = (f(x) \cdot x_1, 0, \dots, 0), \hspace{5pt} x(0) = 0,$$and if we set $g(x_1) := f(x_1,0,\dots,0)$, then $x_1' = g(x_1)x_1$, $x_2 = \cdots = x_n = 0.$ If we define $$G(x_1) := \int \frac{dx_1}{g(x_1)x_1},$$ then, heuristically, $$x_1 = G^{-1}(t),$$ and both of the trajectories are contained in the $\{ x_2 = \cdots = x_n = 0\}$-part of the chart.
However, I don't think this is a difficult question and I'd like to see a formal argument. I suspect that the only reason that I'm unable to resolve this is because of my (very) rusty knowledge of the theory of ODE's.
You'll need some additional smoothness properties of $X$ (and $f$), but with those in hand, the answer is yes.
The reason for this answer is the uniqueness part of the existence and uniqueness theorem for solutions of ODE's (for example look at Theorem 2 here). That theorem says that if $X$ is $C^1$ then then for each initial condition $(x,v)$ ($x \in M$, $v \in T_x M$) the solution is unique over some open interval.
Now the idea is to observe that any solution for $fX$ with initial condition $(x,v)$ can be reparameterized to give a solution for $X$ with the same initial condition $(x,v)$. To put this another way, by a change of variables one converts a solution for $fX$ into a solution for $X$. The key point is that a change of variable does not affect the image set of the solution, i.e. it does not affect the trajectory as a subset of $M$.