Linearised diffeomorphisms on arbitrary gravitational background Part 2

This question is a follow on from my recent post here, in the sense that I will use the notation introduced there. In that post, I considered infinitesimal diffeomorphisms of a metric $g_{munu}$ expanded about some arbitrary fixed background metric $bar{g}_{munu}$ via
begin{align}
g_{munu}=bar{g}_{munu}+h_{munu}~. tag{1}
end{align}
I found that to first order in the diffeomorphism parameter $xi^{mu}$ we have
begin{align}
delta_{xi}g_{munu} = delta_{xi} bar{g}_{munu}+delta_{xi} h_{munu}= bar{nabla}_{mu}xi_{nu}+ bar{nabla}_{nu}xi_{mu} +mathcal{L}_{xi}h_{munu}~,tag{2}
end{align}
where $bar{nabla}_{mu}$ is the (torsion-free, metric compatible) covariant derivative for the background $bar{g}_{munu}$, I have used $bar{g}_{munu}$ to raise/lower indices and $mathcal{L}_{xi}$ is the Lie derivative along $xi^{mu}$,
begin{align}
mathcal{L}_{xi}h_{munu}=xi^{rho}partial_{rho}h_{munu}+partial_{mu}xi^{rho}h_{rhonu}+partial_{nu}xi^{rho}h_{murho}=xi^{rho}bar{nabla}_{rho}h_{munu}+bar{nabla}_{mu}xi^{rho}h_{rhonu}+bar{nabla}_{nu}xi^{rho}h_{murho}~. tag{3}
end{align}

It seems to me that there are two ways of interpreting the transformation rule (2):

1. Passive interpretation: Both the background and perturbation $h_{munu}$ transform under the linearised diffeomorphism according to
   begin{align}
   delta_{xi}bar{g}_{munu}=bar{nabla}_{mu}xi_{nu}+ bar{nabla}_{nu}xi_{mu} ~, qquad delta_{xi}h_{munu}=mathcal{L}_{xi}h_{munu}~. tag{4}
   end{align}
2. Active interpretation: The background does not transform at all, and the whole burden of transformation must fall onto the perturbation,
   begin{align}
   delta_{xi}bar{g}_{munu}=0 ~, qquad delta_{xi}h_{munu}=bar{nabla}_{mu}xi_{nu}+ bar{nabla}_{nu}xi_{mu}+mathcal{L}_{xi}h_{munu}~. tag{5}
   end{align}

I have two questions, one corresponding to each interpretation. To better articulate them, it will help to have in mind some concrete gravitational model, and for this I’ll pick Einstein-Hilbert (EH) gravity,
begin{align}
S_{text{EH}}[g]=int text{d}^dx sqrt{-g}Big( R[g]-2LambdaBig)~. tag{6}
end{align}
Using (1) this may be expanded to quadratic order in $h_{munu}$ according to
begin{align}
S_{text{EH}}[g]=S_{text{EH}}[bar{g}]+{}^{(1)}S_{text{EH}}[bar{g},h]+{}^{(2)}S_{text{EH}}[bar{g},h]~.tag{7}
end{align}
Here the zeroth order term $S_{text{EH}}[bar{g}]$ is the EH action for the background metric, whilst to first order in $h_{munu}$ we have
begin{align}
{}^{(1)}S_{text{EH}}[bar{g},h]propto int text{d}^dx sqrt{-bar{g}} h^{munu}bar{G}_{munu}[bar{g}] tag{8}
end{align}
where $bar{G}_{munu}[bar{g}]$ is the background Einstein tensor. To second order in $h_{munu}$ we have
begin{align}
{}^{(2)}S_{text{EH}}[bar{g},h]propto inttext{d}^dx sqrt{-bar{g}} h^{munu} mathcal{A}_{munu,rhosigma}[bar{g}]h^{rhosigma}~, tag{9}
end{align}
for some linear two derivative kinetic operator $mathcal{A}$ dependent only on $bar{g}_{munu}$.

First Question

In some sense the passive interpretation makes sense to me and in some ways it does not.

For example, it expresses the fact that we have diffeomorphism covariance of the background, in the sense that we could consider the right hand side of (7) in its own right, and it should also be invariant under diffeomorphisms. This may be what some refer to as background independence.

On the other hand, when using the expansion (1), people often say that they are expanding around a fixed background. How is this consistent with the passive interpretation above? In what sense is the background $bar{g}_{munu}$ fixed if it is defined modulo the transformations $delta_{xi} bar{g}_{munu}$? What if the symmetry group was larger, like in conformal gravity where there is also Weyl symmetry. Is performing a Weyl transformation of the background metric not in contention with that statement that the background is fixed?

Second Question

The active interpretation seems to be the origin of the statement that $h_{munu}$ is defined modulo the gauge transformations
begin{align}
hat{delta}_{xi} h_{munu}:=bar{nabla}_{mu}xi_{nu}+ bar{nabla}_{nu}xi_{mu}~. tag{10}
end{align}
I have often read that when linearising an action, as done in (7), one must restrict to background configurations which satisfy the equations of motion. In the case of the EH action this means restricting to Einstein backgrounds.

To me though, it seems that the expansion (7) is valid for arbitrary backgrounds and if one includes the Lie derivative term in (5), then the ‘gauge invariance’ holds to first order in $xi^{mu}$. In particular, in the gauge variation there will be a non-zero term from the first order sector which communicates with the second order sector. The corresponding invariance condition is of the form
begin{align}
delta_{xi}S_{text{EH}}[g]|_{mathcal{O}(h^2)}= gamma_1 color{red}{mathcal{L}_{xi}h^{munu}bar{G}_{munu}[bar{g}]} + 2gamma_2hat{delta}_{xi}h^{munu}mathcal{A}_{munu,rhosigma}[bar{g}]h^{rhosigma} ~, tag{11}
end{align}
for some unimportant constants $gamma_1$ and $gamma_2$.

If one doesn’t include the important Lie derivative term in the variation of $h_{munu}$ in (5), then the term in red on the RHS of (11) will not be there. Hence on an arbitrary background, the quadratic sector ${}^{(2)}S_{text{EH}}[bar{g},h]$ cannot be gauge invariant on its own under the gauge variation defined by (10). However, if one restricts the background to be Einstein then the red term vanishes identically, and in this sense the quadratic action (9) is gauge invariant under (10).

Is this what people mean when they say one must linearise the action (6) about an Einstein background?