= Solution
The same invariant interval can be written as $g_{\mu\nu}dx^\mu dx^\nu=\widetilde g_{\alpha\beta}d\widetilde x^\alpha d\widetilde x^\beta$. Applying the <chain rule> to the differentials proves the \b[metric transformation law]
$$
\boxed{g_{\mu\nu}(x)=
\frac{\partial\widetilde x^\alpha}{\partial x^\mu}
\frac{\partial\widetilde x^\beta}{\partial x^\nu}
\widetilde g_{\alpha\beta}(\widetilde x).}
$$
For first-order <scalar cosmological perturbations> and $\xi^\mu=(T,\partial^iL)$, compare perturbations at the same background coordinate label. Expanding the Jacobians and the shifted background gives the <linear metric gauge-transformation law>
$$
\widetilde{\delta g}_{\mu\nu}
=\delta g_{\mu\nu}-\mathcal L_\xi\bar g_{\mu\nu}.
$$
For the background $\bar g_{00}=-a^2$, $\bar g_{ij}=a^2\delta_{ij}$, the <Lie derivative> components are
$$
(\mathcal L_\xi\bar g)_{00}=-2a^2(T'+\mathcal HT),\qquad
(\mathcal L_\xi\bar g)_{0i}=a^2\partial_i(L'-T),
$$
$$
(\mathcal L_\xi\bar g)_{ij}=2a^2\mathcal HT\delta_{ij}
+2a^2\partial_i\partial_jL,
\qquad \mathcal H=\frac{a'}a.
$$
Here primes denote <conformal time> and $\mathcal H$ is the <conformal Hubble parameter>. These expressions establish all four alternatives, including any chosen pair.
The $00$ component is $\delta g_{00}=-2a^2A$, immediately giving the <lapse function> perturbation transformation. The $0i$ component is $a^2\partial_iB$, giving the <shift vector> scalar-potential transformation. The trace of the spatial perturbation is $6a^2C$, so tracing the spatial <Lie derivative> gives the transformation of $C$. Its trace-free scalar part is $2a^2(\partial_i\partial_j-\delta_{ij}\nabla^2/3)E$, giving the transformation of $E$. Thus
$$
\boxed{\begin{aligned}
\widetilde A&=A-T'-\mathcal HT,\\
\widetilde B&=B+T-L',\\
\widetilde C&=C-\mathcal HT-\frac13\nabla^2L,\\
\widetilde E&=E-L.
\end{aligned}}
$$
As usual, identifying scalar potentials from their derivatives uses the standard boundary conditions, or nonzero Fourier modes, to remove homogeneous ambiguities.
At first order, <tensor cosmological perturbations> are spatial <transverse-traceless tensors>. The coordinate-generated spatial perturbation consists of a trace term and symmetrized derivatives of the displacement. In <Fourier space>, the latter terms carry a factor $k_i$ or $k_j$. The <transverse-traceless projector> removes these longitudinal terms and the trace, including the derivative of a transverse vector displacement. Therefore \b[the tensor perturbation is gauge invariant at linear order around the homogeneous background]. This is a first-order statement, not a claim of automatic invariance at arbitrary perturbative order.
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