= Solution
Let $\delta h_{\mu\nu}=\partial_\mu V_\nu+\partial_\nu V_\mu$. Substitution into the <linearized Levi-Civita connection> cancels four terms and leaves
$$
\delta\Gamma^\mu{}_{\nu\rho}=\partial_\nu\partial_\rho V^\mu.
$$
To first order in the <metric perturbation>, the products of <Christoffel symbols> are quadratic and may be discarded. Therefore the change in the <Riemann curvature tensor> is
$$
\delta R^\mu{}_{\nu\rho\sigma}
=\partial_\rho\delta\Gamma^\mu{}_{\sigma\nu}-\partial_\sigma\delta\Gamma^\mu{}_{\rho\nu}
=\partial_\rho\partial_\sigma\partial_\nu V^\mu-\partial_\sigma\partial_\rho\partial_\nu V^\mu=0.
$$
This is the <gauge invariance of the linearized Riemann tensor>. Lowering the first index with the <Minkowski metric> does not change the conclusion; the change from using the full <metric tensor> instead is of second order because the background curvature is zero. Hence
$$
\boxed{\delta R^{(1)}_{\mu\nu\rho\sigma}=0.}
$$
The result is \b[invariance at linear order about flat spacetime], not componentwise invariance of arbitrary nonlinear curvature under finite coordinate changes. Directly in the four-term formula, the eight third derivatives of $V$ cancel in pairs by commuting partial derivatives.
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