Gauge dependence of the Einstein tensor and the Riemann/Ricci curvature tensors in non-linear general relativity The Einstein field equations are given by (with assuming $\Lambda = 0$),
$$
R_{ab} - \frac{1}{2} R g_{ab} = \kappa T_{ab}.
$$
The principle of general covariance states that the form of these equations are invariant under diffeomorphisms. These equations can be linearized, by introducing
$$
g_{ab} = \eta_{ab} + h_{ab}
$$
and letting $h_{ab}$ be small. The coordinate change $x'^\mu = x^\mu + \xi^\mu(x^\mu)$ changes the perturbation metric $h$ as
$$
h'_{ab} = h_{ab} + \mathcal{L}_\xi\eta_{ab} =  h_{ab} + \partial_a \xi_b + \partial_b \xi_a.
$$
In the linear theory, which can be readily checked, the Riemann tensor is invariant under the transformation mentioned above, which also means that the energy-momentum tensor $T_{ab}$ is invariant to first order in $h$.
So, now comes my question:
In the full non-linear theory, the gauge freedom is given by general changes of coordinate systems. My understanding goes as follows; the Ricci/Riemann tensors express curvature, which is an intrinsic property of the manifold, and as this curvature is invariant of coordinate changes (or diffeomorphisms?). The "curvature" property of these tensors have to be invariant. Still, they are tensors, and should therefore act as tensors under rotations, boosts, etc., so they can't be invariant either.
I have only seen the gauge transformation properties discussed in the linear theory, and I am having problems understanding what it "means" in this setting to posses an intrinsic property which should be invariant of coordinate system. All of $R_{abcd}$, $R_{ab}$ and $T_{ab}$ posesses some physcial meaningful value, and should therefore have some "invariant part". They should possess some equivalence classes that are different, and not possible to transform into one another by coordinate changes. What does this mean, and what is this "invariant part"? Also, is there a fundamental difference between a gauge transformation/diffeomorphism and a global transformation (such as a rotation/boost)?
 A: Einstein's field equations (EFE) in nonlinear form are not a gauge theory. Attempts have been made to formulate it as a gauge theory, but this requires the use of the tetrad's formalism so that the EFE can be formulated in a coordinate-independent way. As tentative gauge group the local Poincare' -invariance is required. However, as far as I know the EFE cannot be formulated completely gauge -invariant under this group. Actually different attempts exist, but the formulation of the EFE as gauge theory is out of scope here. On the other hand the EFE are coordinate-covariant which can be expressed simply by the fact that the form of the EFEs remains the same under any (at least differentiable) coordinate transformations (this is what sometimes is called invariance under diffeomorphisms)
$$\overline{x^i}=\overline{x^i}(x^i)$$ 
and each of its tensors $R_{ik}$, $g_{ik}$ and $T_{ik}$ transforms in the same way (I take  $R_{ik}$ as an example):
$$R_\overline{ik} = \frac{\partial\overline{x^i}}{\partial x^l}\frac{\partial\overline{x^k}}{\partial x^m} R_{lm}$$
As all tensors transform in the same way, it is guaranteed that the EFE don't change their form. Another detail: whatever coordinate transformation is carried out, a tensor is only zero if it is zero in all coordinate systems. Actually, the coordinate change $x'^i = x^i+\xi^i(x)$ is a linearization of the more general coordinate transformations $\overline{x^i}=\overline{x^i}(x)$, so the gauge transformation of the tensor $h_{ik}$ in your post just expresses the more general coordinate covariance. 
