The meaning of diffeomorphism invariance in general relativity

Solution 1:

This may not be the answer you are asking for, but regardless: if you pull back the metric with $f$, then, by definition $$\tilde g(f(p))(f_{*,p} X, f_{*,p}Y)=f^* g(p)(X,Y)$$ which means, (again) by definition, that $f$ is an isometry between $(M, g)$ and $(M,\tilde{g})$. This implies that you cannot geometrically distinguish quantities which are defined through the metric in these manifolds, which then implies (without further computation!) that your curvature related equations are fulfilled in $(M,g)$ if and only if that's true for $(M, \tilde{g})$.

This is assuming you are using the Levi Civita connection, which is what you have claimed. Of course you have to use the Levi Civita connection with respect to $\tilde{g}$ in $(M, \tilde{g})$ and the one with respect to $g$ in $(M,g)$.

The less formal reasoning is that the mathematical description of a physical process must not depend on the mathematical representation, which is, by the way, also a paradigm in (differential) geometry. The way the earth is moving around the sun does not (must not) change just because you switch from, say, cartesian to cylindrical coordinates.

(I can easily believe that you have trouble to verify the computations, especially in the coordinate free formalism, as these calculations offer many opportunities for errors. I don't know, though, where your mistake is...).