TITLE: Killing vector fields of a conformally flat Riemannian metric
QUESTION [5 upvotes]: 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!

REPLY [4 votes]: Hint
Assume that $h=e^fg$ is a conformal change of $g$ on a manifold $M$. Let $X$ be a vector field on $M$. Then
$${\cal{L}}_{X}h={\cal{L}}_{X}(e^fg)\\
={\cal{L}}_{X}(e^f)g+e^f{\cal{L}}_{X}g\\
=e^fX(f)g+e^f{\cal{L}}_{X}g$$
where ${\cal{L}}_{X}$ denotes the Lie derivative in direction of $X$ on $M$. This leads to say "Killing fields of both metrics are the same if and only if $f$ is constant along the flow lines of these vector fields"