= Solution
Work to first order in a <metric perturbation> $g_{ab}=\eta_{ab}+h_{ab}$, with $\eta=\operatorname{diag}(1,-1,-1,-1)$, raising perturbation indices with $\eta$ and discarding quadratic terms. The <Christoffel symbols> are
$$
\Gamma^{a(1)}{}_{bc}=\frac12\eta^{ad}(\partial_bh_{dc}+\partial_ch_{db}-\partial_dh_{bc}).
$$
Using the printed curvature convention gives the <linearized Riemann curvature operator>
$$
R^{(1)}_{abcd}=\frac12(\partial_b\partial_dh_{ac}+\partial_a\partial_ch_{bd}-\partial_a\partial_dh_{bc}-\partial_b\partial_ch_{ad}).
$$
For definiteness contract $R_{bd}=R^a{}_{bad}$. The corresponding first-order <Ricci tensor> is
$$
R^{(1)}_{ab}=\frac12(\partial_a\partial_bh+\Box h_{ab}-\partial_a\partial^ch_{cb}-\partial_b\partial^ch_{ca}),\qquad\Box=\partial_t^2-\Delta.
$$
This is the reversed sign convention specified in <curvature-sign convention in Killing derivative identities>; the vacuum equation is unaffected by an overall curvature sign.
An infinitesimal passive coordinate change $x'^a=x^a+\xi^a$ changes the perturbation by $\delta h_{ab}=-\partial_a\xi_b-\partial_b\xi_a$. Substitution into $R^{(1)}$ gives zero change, since partial derivatives commute. Equivalently the change in curvature is the Lie derivative of the background curvature, which vanishes in <Minkowski spacetime>. Thus linearized curvature, and the linearized <Weyl tensor> in vacuum, are gauge-invariant tidal observables, whereas individual components of $h$ are generally gauge dependent.
The <trace-reversed metric perturbation> $\bar h_{ab}=h_{ab}-\eta_{ab}h/2$ transforms by $\delta\bar h_{ab}=-\partial_a\xi_b-\partial_b\xi_a+\eta_{ab}\partial_c\xi^c$. Its divergence changes by $-\Box\xi_b$. Solving $\Box\xi_b=\partial^a\bar h_{ab}$ therefore imposes <Lorenz gauge in linearized gravity>, $\partial^a\bar h_{ab}=0$. With the stated sign convention the <Einstein tensor> is
$$
G^{(1)}_{ab}=\frac12[\Box\bar h_{ab}-\partial_a\partial^c\bar h_{cb}-\partial_b\partial^c\bar h_{ca}+\eta_{ab}\partial^c\partial^d\bar h_{cd}],
$$
so the linearized vacuum field equation becomes $\boxed{\Box\bar h_{ab}=0}$. The <residual gauge symmetry of linearized gravity> has $\Box\xi_a=0$. This makes a hyperbolic evolution problem after the gauge condition and its initial consistency constraints have been imposed.
To separate actual gauge invariants from a gauge choice, use <gauge-invariant Minkowski perturbations>. For spatial indices contracted with $\delta_{ij}$, decompose
$$
h_{00}=2A,\quad h_{0i}=\partial_iB+B_i,\quad h_{ij}=2C\delta_{ij}+2(\partial_i\partial_j-\delta_{ij}\Delta/3)E+\partial_iE_j+\partial_jE_i+h_{ij}^{\rm TT},
$$
where $\partial_iB_i=\partial_iE_i=0$, $\partial_ih_{ij}^{\rm TT}=0$ and $h_{ii}^{\rm TT}=0$. Spatial falloff or Fourier-mode conventions fix the decomposition. Write $\xi^0=\tau$ and $\xi^i=\partial_iL+L_i$ with $\partial_iL_i=0$. Direct substitution in the gauge change gives
$$
\delta A=-\dot\tau,\quad\delta B=\dot L-\tau,\quad\delta C=\Delta L/3,\quad\delta E=L,\quad\delta B_i=\dot L_i,\quad\delta E_i=L_i,\quad\delta h_{ij}^{\rm TT}=0.
$$
Hence the invariant combinations are
$$
\boxed{\Phi=A-\dot B+\ddot E,\qquad\Psi=C-\Delta E/3,\qquad V_i=B_i-\dot E_i,\qquad h_{ij}^{\rm TT}.}
$$
A representative with $E=E_i=B=0$ has $h_{00}=2\Phi$, $h_{0i}=V_i$ and $h_{ij}=2\Psi\delta_{ij}+h_{ij}^{\rm TT}$. The explicit <vacuum constraints for gauge-invariant Minkowski perturbations> follow from
$$
G^{(1)}_{00}=-2\Delta\Psi,\qquad G^{(1)}_{0i}=-2\partial_i\dot\Psi-\tfrac12\Delta V_i,
$$
$$
G^{(1)}_{ij}=(\partial_i\partial_j-\delta_{ij}\Delta)(\Phi-\Psi)-2\delta_{ij}\ddot\Psi-\tfrac12(\partial_i\dot V_j+\partial_j\dot V_i)+\tfrac12\Box h_{ij}^{\rm TT}.
$$
For a nonzero spatial Fourier mode, or regular decaying fields on all space, $G_{00}=0$ forces $\Psi=0$, then $G_{0i}=0$ forces $V_i=0$, and the spatial trace gives $\Delta\Phi=0$, hence $\Phi=0$. Only $\Box h_{ij}^{\rm TT}=0$ remains. This removal of the scalar and vector sectors does not exclude singular Coulomb perturbations outside a source, boundary-driven harmonic potentials or spatially homogeneous modes; those require separate boundary and gauge conventions.
For a propagating plane mode $h\propto e^{ik_ax^a}$, the wave equation gives $k^ak_a=0$. A transverse-traceless spatial tensor perpendicular to a fixed propagation direction has two independent components. For propagation along $z$, define the physical strain $H_{ij}=-h_{ij}^{\rm TT}$ in this mostly-minus convention, so the spatial line element is $(\delta_{ij}+H_{ij})dx^idx^j$. It has the form
$$
H_{ij}(t,z)=\begin{pmatrix}H_+(t-z)&H_\times(t-z)&0\\H_\times(t-z)&-H_+(t-z)&0\\0&0&0\end{pmatrix}.
$$
These are the two <gravitational wave polarizations>, <plus polarization> and <cross polarization>. A rotation in the transverse plane rotates the polarization pair through twice the angle; cross is plus rotated through $45$ degrees. Lorenz constraints and residual gauge freedom give the same count: ten symmetric metric components minus four constraints and four gauge freedoms leave two radiative degrees of freedom.
In <transverse-traceless gauge>, particles initially at rest keep fixed spatial coordinates since $\Gamma^a{}_{00}=0$, but their measured separation changes. For a short separation of length $L$ in unit direction $n$, $\delta L/L=H_{ij}n^in^j/2$. The gauge-invariant curvature gives the corresponding <geodesic deviation>:
$$
R^{i(1)}{}_{00j}=\tfrac12\ddot h_{ij}^{\rm TT}=-\tfrac12\ddot H_{ij},\qquad\ddot\xi^i=-R^{i(1)}{}_{00j}\xi^j=\tfrac12\ddot H_{ij}\xi^j.
$$
Here the last equation is in a freely falling orthonormal frame, with the printed reversed curvature convention. Coordinate separation and physical separation must not be confused. The linearized vacuum equations describe propagation and tidal measurements; wave energy transfer and an effective wave <stress-energy tensor> arise at quadratic order, beyond the linear approximation.
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