= Solution
{c}
Use the active sign convention for the <linearized coordinate gauge transformation>, $h_{\mu\nu}\mapsto h_{\mu\nu}+\partial_\mu\xi_\nu+\partial_\nu\xi_\mu$. The paper's prescribed generator has precisely this sign; using the passive convention would require its negative. Direct <partial differentiation> gives
$$
\partial_\nu\xi_\mu=\frac14\left[
\tilde h''_{\kappa\sigma}x^\kappa x^\sigma n_\mu n_\nu
+2\tilde h'_{\nu\kappa}x^\kappa n_\mu
-2\tilde h'_{\mu\kappa}x^\kappa n_\nu-2\tilde h_{\mu\nu}\right].
$$
The middle terms cancel when the indices are symmetrized. Therefore
$$
\partial_\mu\xi_\nu+\partial_\nu\xi_\mu
=\frac12\tilde h''_{\kappa\sigma}x^\kappa x^\sigma n_\mu n_\nu-\tilde h_{\mu\nu},
\qquad
\boxed{\hat h_{\mu\nu}=\phi n_\mu n_\nu,\quad
\phi=\frac12\tilde h''_{\kappa\sigma}x^\kappa x^\sigma.}
$$
To test both equations (3), put $H(u)=\tilde h(u)=\eta^{\kappa\sigma}\tilde h_{\kappa\sigma}(u)$, the <trace> of the original <metric perturbation>. Since $n$ is a <null vector>, $\hat h=0$ and its <trace-reversed metric perturbation> equals $\hat h$. Differentiating the result of part (a) gives
$$
n\cdot\partial\phi=n^\kappa\tilde h''_{\kappa\sigma}x^\sigma
=\frac12uH'',
\qquad
\Box\phi=H''+2n^\kappa\tilde h'''_{\kappa\sigma}x^\sigma
=H''+uH'''.
$$
In the second identity, the term with $n^2$ is zero. Thus
$$
\boxed{\partial^\mu\overline{\hat h}_{\mu\nu}=\frac12uH''n_\nu,\qquad
\Box\hat h_{\mu\nu}=(H''+uH''')n_\mu n_\nu.}
$$
\b[In general $\hat h$ does not satisfy (3).] On an open interval of $u$, both equations hold exactly when $H''=0$ there: the first equation forces this for $u\ne0$, and smoothness gives it at $u=0$ as well. A <linearized coordinate gauge transformation> preserves the <linearized vacuum Einstein equations>, but only a <residual gauge symmetry of linearized gravity> preserves the imposed <Lorenz gauge in linearized gravity>.
For an explicit failure, use the <null coordinates> of part (c) and take only $\tilde h_{uv}=\tilde h_{vu}=u^2/4$ nonzero. Then $H=u^2$, the original field obeys (3), but $\phi=uv/2$ gives $\partial^\mu\overline{\hat h}_{\mu u}=u$ and $\Box\hat h_{uu}=2$.
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