Let $f: \mathbb{R}^n \to \mathbb{R}$ be a smooth function and let's consider the conformally flat Riemannian metric $g = e^f \delta_{ij} dx^idx^j$ on $\mathbb{R}^n$.

Is it true that the Killing vector fields of the manifold $(\mathbb{R}^n, g)$ are exactly the Killing fields of the Euclidean space such that $f$ is constant along their flow? For sure if $K$ is an Euclidean Killing vector field with such property, then it must be a Killing field also for the manifold $(\mathbb{R}^n, g)$, but I'm not completely sure that those are exactly all the possible Killing fields of $(\mathbb{R}^n, g)$. I checked some simple cases and it seems to be the case.

Is it true in general? Or is it possible to find a counterexample where $(\mathbb{R}^n, g)$ has some symmetries that do not arise from symmetries of $f$?

Any help will be very appreciated! Thanks a lot!