= Solution
Work consistently to first order in the <metric perturbation>, with $\partial_\mu\xi^\nu$ of the same small order. Under the passive change $x'^\mu=x^\mu+\xi^\mu(x)$,
$$
\frac{\partial x^\alpha}{\partial x'^\mu}
=\delta^\alpha_\mu-\partial_\mu\xi^\alpha+O(h^2).
$$
Applying this to the full <metric tensor> and writing $g'_{\mu\nu}=\eta_{\mu\nu}+h'_{\mu\nu}$ with the same fixed numerical <Minkowski metric> gives
$$
\boxed{
h'_{\mu\nu}(x')=h_{\mu\nu}(x)
-\partial_\mu\xi_\nu-\partial_\nu\xi_\mu+O(h^2).
}
$$
Indices on $\xi$ are lowered with $\eta$. This is the gauge transformation of the perturbation relative to a fixed background. If the background were transformed as a tensor as well, the difference of those two fully transformed metrics would instead obey the homogeneous tensor transformation law. The inhomogeneous terms arise from retaining the fixed-background identification used in <linearized gravity>.
The trace changes by $h'=h-2\partial_\alpha\xi^\alpha$, so the <trace-reversed metric perturbation> changes as
$$
\bar h'_{\mu\nu}
=\bar h_{\mu\nu}
-\partial_\mu\xi_\nu-\partial_\nu\xi_\mu
+\eta_{\mu\nu}\partial_\alpha\xi^\alpha.
$$
Let $F_\nu=\partial^\mu\bar h_{\mu\nu}$. Commuting <partial derivatives> gives
$$
\partial^\mu\bar h'_{\mu\nu}
=F_\nu-\Box\xi_\nu,\qquad
\Box=\eta^{\alpha\beta}\partial_\alpha\partial_\beta.
$$
Thus choose a solution of the inhomogeneous <wave equation>
$$
\boxed{\Box\xi_\nu=F_\nu.}
$$
Such a solution exists locally with suitable initial data, and can be selected with appropriate wave-equation <boundary conditions>. This imposes the <Lorenz gauge in linearized gravity>, $\partial^\mu\bar h'_{\mu\nu}=0$. All the divergence terms in the supplied vacuum equation then vanish, leaving
$$
\boxed{\Box\bar h'_{\mu\nu}=0.}
$$
The remaining freedom $\Box\xi_\nu=0$ is the <residual gauge symmetry of linearized gravity>.
Using the old coordinates in these derivatives is consistent at the retained order. Indeed
$$
\partial'_\mu
=\partial_\mu-(\partial_\mu\xi^\alpha)\partial_\alpha+O(h^2).
$$
Acting on a first-order perturbation, the correction is a product of two first-order quantities and is discarded. Similarly replacing the argument $x'$ by $x$ changes $h'$ by $O(h^2)$ under the same perturbative ordering. This does not assert an exact equality of the two coordinate differential operators; it is the linear approximation about the fixed flat background.
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