I want to show that, if $(M,\mathrm{g})$ is a Riemannian manifold, $\nabla$ is the covariant derivative from the Levi-Civita connection, and $f:M\to M$ is an isometry, then $f_*(\nabla_XY)=\nabla_{f_*X}f_*Y$. I can demonstrate that this works if I resort to coordinate expressions, but I don't want to do that.
At first, I thought to use the fact that $\nabla$ is metric, so that $$\nabla_Zf^*\mathrm{g}(X,Y)=\mathrm{g}(f_*(\nabla_ZX),f_*Y)+\mathrm{g}(f_*X,f_*(\nabla_ZY))=\mathrm{g}(\nabla_ZX,Y)+\mathrm{g}(X,\nabla_ZY),$$ but I can't seem to get anywhere with it.
Any hints as to how I might approach this?
Assume that $f\colon M\to M$ is a diffeomorphism satisfying $f^*g = g$.
It's not enough just to use the fact that $\nabla$ is a metric connection; you also have to use the fact that it's torsion-free. The idea of the proof is to define $\widetilde \nabla\colon \mathfrak X(M)\times \mathfrak X(M) \to \mathfrak X(M)$ by $$ \widetilde \nabla_XY = (f^{-1})_*(\nabla_{f_*X}{f_*Y}). $$ Then you can show that $\widetilde\nabla$ is a connection on $M$ that is compatible with $g$ and torsion-free. By the uniqueness of the Riemannian connection, $\widetilde \nabla$ must be equal to $\nabla$.