= Solution
Use a dot for the coordinate-time derivative and let $\bar N$ be the homogeneous background <lapse function>. The passive linear transformation is $\delta\widetilde g_{\mu\nu}=\delta g_{\mu\nu}-\mathcal L_\xi\bar g_{\mu\nu}$. Its time-time and time-space components yield the <scalar gauge transformations with a background lapse>:
$$
\widetilde\Psi=\Psi-\dot\xi^0-\frac{\dot{\bar N}}{\bar N}\xi^0,\qquad \widetilde B=B+\frac{\bar N^2}{a^2}\xi^0-\dot\lambda.
$$
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. Hence
$$
\partial_t(\bar N\xi^0)=0,\qquad\dot\lambda=\frac{\bar N^2}{a^2}\xi^0.
$$
Integrating gives the <residual synchronous-gauge freedom>
$$
\boxed{\xi^0=\frac{C(\mathbf x)}{\bar N(t)},\qquad\lambda=C(\mathbf x)\int^t\frac{\bar N(t')}{a(t')^2}\,dt'+D(\mathbf x).}
$$
The 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 $\widetilde\Phi=\Phi+(\dot a/a)\xi^0$ and $\widetilde E=E+\lambda$. 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>.
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