 Solution

Use a dot for the coordinate-time derivative and let $\bar{N}$ be the homogeneous background lapse function. The passive linear transformation is $\delta {\tilde{g}}_{\mu \nu }=\delta g_{\mu \nu }-{\mathcal{L}}_{\xi }{\bar{g}}_{\mu \nu }$. Its time-time and time-space components yield the scalar gauge transformations with a background lapse:For example, ${\bar{g}}_{00}=-{\bar{N}}^2$ gives $-({\mathcal{L}}_{\xi }\bar{g}{)}_{00}=2\bar{N}\dot{\bar{N}}{\xi }^0+2{\bar{N}}^2{\dot{\xi }}^0$; ${\bar{g}}_{0i}=0$ gives $-({\mathcal{L}}_{\xi }\bar{g}{)}_{0i}={\bar{N}}^2{\partial }_i{\xi }^0-a^2{\partial }_i\dot{\lambda }$. These establish the signs directly.

To preserve synchronous gauge in cosmology, both transformed quantities must remain zero. HenceIntegrating gives the residual synchronous-gauge freedomThe integration lower limit can be absorbed into $D$. Spatially homogeneous additions to $\lambda$ generate no spatial displacement, so they are physically irrelevant to this scalar parametrization.

The spatial transformation also gives $\tilde{\Phi }=\Phi +(\dot{a}/a){\xi }^0$ and $\tilde{E}=E+\lambda$. Thus fixing the lapse and shift perturbations does not fix the entire coordinate system; the arbitrary functions $C,D$ can still change the remaining scalar cosmological perturbations.