Let $X$ and $Y$ be Killing fields on a Riemannian manifold, with $[X,Y]=0$. Is then the total covariant derivative $\nabla Y$ invariant under the flow of $X$? If so, how do you prove it? I've seen a paper where I think they use this result, and I have tried to prove it myself. Mainly, I've just written out the expression for the Lie derivative $\mathcal{L}_X(\nabla Y)$ (in order to show that it is zero): for any vector field $Z$, we have
$$(\mathcal{L}_X(\nabla Y))(Z)=[X,\nabla_Z Y]-\nabla_{[X,Z]}Y.$$
I have had no success though, because I have no idea where to go from here. Any ideas?
I don't think you even need $Y$ to be Killing here. Write $$\mathcal{L}_X(\nabla Y) = (\mathcal{L}_X\nabla)Y + \nabla(\mathcal{L}_XY)$$and note that since $X$ is Killing, then $\mathcal{L}_X\nabla = 0$ (the flow of $X$ consists of isometries, which also preserve the connection), while your additional assumption is just $\mathcal{L}_XY = [X,Y]=0$.
The Lie derivative of the connection is exactly what you think: $$(\mathcal{L}_X\nabla)_YZ = \mathcal{L}_X(\nabla_YZ) - \nabla_{\mathcal{L}_XY}Z - \nabla_Y(\mathcal{L}_XZ).$$