= Solution
Retain the homogeneous <lapse function> $\bar N(t)$, and define the physical <Hubble parameter> $H=\dot a/(\bar Na)$. At first order the scalar <metric perturbations> give $\delta g_{00}=-2\bar N^2\Phi$, $\delta g_{0i}=a^2B_{,i}$, and $\delta g_{ij}=-2a^2(\Psi\delta_{ij}+E_{,ij})$. Applying the passive coordinate displacement component by component gives the <scalar gauge transformations with a background lapse>
$$
\boxed{\widetilde\Phi=\Phi-\dot\xi^0-\frac{\dot{\bar N}}{\bar N}\xi^0,\quad
\widetilde B=B-\dot\lambda+\frac{\bar N^2}{a^2}\xi^0,\quad
\widetilde\Psi=\Psi+\frac{\dot a}{a}\xi^0,\quad
\widetilde E=E+\lambda.}
$$
For example the $0i$ transformation is $\delta\widetilde g_{0i}=a^2\partial_iB-a^2\partial_i\dot\lambda+\bar N^2\partial_i\xi^0$. The spatial isotropic part acquires $-2a\dot a\xi^0\delta_{ij}$, explaining the positive sign in the transformation of $\Psi$.
Preserving <synchronous gauge> requires both transformed $\Phi$ and $B$ to remain zero. The first condition is $\partial_t(\bar N\xi^0)=0$, and the second is $\dot\lambda=\bar N^2\xi^0/a^2$. Integrating them gives
$$
\boxed{\xi^0=\frac{C(\mathbf x)}{\bar N},\qquad
\lambda=C(\mathbf x)\int^t\frac{\bar N(t')}{a(t')^2}dt'+D(\mathbf x).}
$$
Thus fixing the <lapse function> and <shift vector> perturbations does not exhaust the scalar coordinate freedom. The functions $C,D$ describe <residual synchronous-gauge freedom>; an integration-origin change in the integral is absorbed into $D$.
To move from <synchronous gauge> to <Newtonian gauge>, choose $\lambda=-E_S$ to set $\widetilde E=0$. Requiring $\widetilde B=0$ then gives
$$
\xi^0=-\frac{a^2}{\bar N^2}\dot E_S.
$$
A scalar <energy density> transforms as $\delta\rho_N=\delta\rho_S-\dot{\bar\rho}\xi^0$. Dividing by the background <energy density>, which changes the contrast only at second order if perturbed in that denominator, yields the <synchronous-to-Newtonian density transformation with a lapse>
$$
\boxed{\delta_N=\delta_S+\frac{a^2\dot E_S}{\bar N^2}\frac{\dot{\bar\rho}}{\bar\rho}
=\delta_S-3(1+w)\frac{a^2H}{\bar N}\dot E_S,
\qquad\delta=\frac{\delta\rho}{\bar\rho}.}
$$
The last form uses background <stress-energy conservation>, $\dot{\bar\rho}=-3\bar NH(\bar\rho+\bar P)$. With <cosmic time> $\bar N=1$ this is $\delta_N=\delta_S-3(1+w)a^2H\dot E_S$; with <conformal time> $\bar N=a$ it is $\delta_N=\delta_S-3(1+w)\mathcal H E_S'$, where $\mathcal H=a'/a$. Keeping these clock conventions distinct is essential to the factors of $a$.
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