Let $f$ be a smooth function and $\nabla_V V = 0$. Prove that $fV \neq 0$ is a killing vector field, if and only if $f$ is a constant and $V$ is killing.
If I assume that $f$ is constant and $V$ is killing, I can easily show that this is satisfied, since $\mathcal{L}_{fV} g = f(\mathcal{L}_V g)-D_{V \otimes df} g = 0$. But showing that $\mathcal{L}_{fV}g=0$ means $V$ is killing and $f$ is constant is a bit more tricky. I suspect that I have to use that $\nabla_V V = 0$. I've tried using some tensor product identities on $V \otimes df$ but it doesn't get me anywhere.
It will be easier to use the characterization of Killing vector fields in terms of covariant derivatives: $V$ is Killing iff $\operatorname{Sym} (\nabla V)^\flat\equiv 0$. Inserting $V$ into one slot of this symmetric 2-tensor field yields an equation that implies $df=0$, and from there it's an easy matter to show that $V$ is Killing.