The Lie derivative of any tensor field $K$ with respect to a vector field $X$ \xi$is by definition $$L_X(K)=-\frac L_\xi(K)=-\frac d{dt}|_{t=0}(\phi_t)_*K,$$ where$\phi_t$is the local flow of$X$. \xi$. Now, if $M$ has a Riemannian metric $g$ and $X$ \xi$is Killing with respect to$g$, each$\phi_t$is a local isometry of$(M,g)$. From the uniqueness of the Levi-Civita connection, it follows that every isometry$\phi$is affine, i.e. $$\phi_*(\nabla_XY)=\nabla_{\phi_*X}\phi_*Y.$$$\phi_*(\nabla_X Y)=\nabla_{\phi_*X}\phi_*Y.$$From here you get immediately \phi_*R=R for the Riemannian curvature, and since the Ricci tensor of R is obtained by a trace:$$Ric(X,Y)=trace(V\mapsto R_{V,X}Y),$$one gets \phi_*Ric=Ric for every isometry \phi. The first formula thus shows that L_X L_\xi Ric=0 for every Killing vector field X.\xi. Edit: Here is another, purely tensorial, proof of the same statement. Let \xi be Killing, in the sense that g(\nabla_X\xi,Y)+g(X,\nabla_Y\xi)=0 for all vector fields X,Y. After taking the covariant derivative wrt some vector field Z, and doing some standard manipulations, one gets the usual Kostant formula:$$\nabla^2_{X,Y}\xi=R_{\xi,X}Y,\qquad\forall X,Y\in C^\infty(TM).$$This is just a rewriting of$$L_\xi(\nabla_XY)=\nabla_{L_\xi X}Y+\nabla_X(L_\xi Y),$$i.e. some sort of Leibniz formula. Applying this formula several times eventually yields the corresponding Leibniz formula for R:$$L_\xi(R_{X,Y}Z)=R_{L_\xi X,Y}Z+R_{X,L_\xi Y}Z+R_{X,Y}(L_\xi Z),$$i.e. L_\xi R=0, and finally L_\xi Ric=0 after taking the trace. Of course, this is just the infinitesimal version of the first proof... 1 The Lie derivative of any tensor field K with respect to a vector field X is by definition$$L_X(K)=-\frac d{dt}|_{t=0}(\phi_t)_*K,$$where \phi_t is the local flow of X. Now, if M has a Riemannian metric g and X is Killing with respect to g, each \phi_t is a local isometry of (M,g). From the uniqueness of the Levi-Civita connection, it follows that every isometry \phi is affine, i.e.$$\phi_*(\nabla_XY)=\nabla_{\phi_*X}\phi_*Y.$$From here you get immediately \phi_*R=R for the Riemannian curvature, and since the Ricci tensor of R is obtained by a trace:$$Ric(X,Y)=trace(V\mapsto R_{V,X}Y),
one gets $\phi_*Ric=Ric$ for every isometry $\phi$. The first formula thus shows that $L_X Ric=0$ for every Killing vector field $X$.