= Solution
Use the plus-sign transformation rule for $h_{\mu\nu}$ stated in this part. A generator $\xi^\mu=b^\mu\sin\Psi$ with constant $b^\mu$ produces the <pure-gauge plane wave in linearized gravity>
$$
\delta\phi^{\mu\nu}=\left[k^\mu b^\nu+k^\nu b^\mu-\eta^{\mu\nu}(k\cdot b)\right]\cos\Psi.
$$
Because $k^2=0$, it obeys both $\Box\xi^\mu=0$ and $k_\mu\delta\epsilon^{\mu\nu}=0$, preserving the imposed <Lorenz gauge in linearized gravity>.
For an explicit example choose $b^\mu=B k^\mu/(2\omega^2)$, with constant small $B$. Since $k\cdot b=0$, this gives
$$
\boxed{\epsilon_{\rm gauge}^{\mu\nu}
=B\begin{pmatrix}1&0&0&1\\0&0&0&0\\0&0&0&0\\1&0&0&1\end{pmatrix},\qquad
\xi_\mu=\frac{B}{2\omega}(1,0,0,-1)_\mu\sin[\omega(t-z)].}
$$
Indeed, direct differentiation gives $\delta h_{\mu\nu}=Bk_\mu k_\nu\cos\Psi/\omega^2$, whose <trace> is zero, so $\delta\phi=\delta h$. This wave-shaped metric disturbance is only a coordinate description of flat spacetime: its <linearized Riemann curvature operator> vanishes. All its second-derivative terms are proportional to four factors of $k$, and the antisymmetric curvature combinations cancel.
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