general relativity - Breaking of diffeomorphism invariance after fixing a background metric

The Lagrangian for the gravitational field in absence of matter is the following

$$L=1/k\int dx^4 \sqrt g R,$$ where $k=\sqrt G$, $g$ is the determinant of the metric and $R$ the Ricci scalar. It's possible to fix a background metric like $\eta_{uv}$ and then study the perturbations $h_{uv}$ around it by $$g_{uv}=\eta_{uv}+kh_{uv}$$ The Lagrangian becomes $$L=L^{0}+kL^{1}+k^{2}L^{2}+.......$$ which can be interpreted as an effective field theory of self-interacting particles called gravitons. Now, given the transformation law of $h_{uv}$, how is it possible to say the entire Lagrangian is invariant, order by order, under local diffeomorphisms? Of course the symmetry is still there, but I was wondering if there is some kind of Spontaneus Symmetry Breaking associated with the perturbation field $h_{uv}$ and the diffeomorphisms group. The procedure resemble the SSB for the Higgs Boson, where the Lagrangian is $$L=\partial_{u}\phi\partial^{u}\phi - m^{2}\phi^{2}+\lambda\phi^{4}$$ This Lagrangian is invariant under parity in $\phi$, but after the redefinition around the vacuum $v$, the minimum of the potential, you deal with $\phi=v+\delta\phi$ and the Lagrangian in δϕ is no more parity invariant. Does this happen in the previous example after fixing a background?