= Solution
For the convention in the question, a finite <Yang-Mills gauge transformation> acts covariantly on the field strength:
$$
F_{\mu\nu}\mapsto F'_{\mu\nu}=UF_{\mu\nu}U^\dagger
$$
or with $U$ and $U^\dagger$ exchanged if the opposite convention is used for $D_\mu$. Cyclicity of the <matrix trace> gives
$$
\operatorname{Tr}(F'_{\mu\nu}F'^{\mu\nu})
=\operatorname{Tr}(UF_{\mu\nu}F^{\mu\nu}U^\dagger)
=\operatorname{Tr}(F_{\mu\nu}F^{\mu\nu}),
$$
so the <Yang-Mills theory> Lagrangian is gauge invariant.
The quadratic gauge-field operator has zero directions $A_\mu\sim A_\mu+D_\mu\alpha$ along each <gauge orbit>. It therefore has no inverse on the full field space. <Gauge fixing> removes this degeneracy and produces a propagator, while the <Faddeev-Popov determinant> accounts for the corresponding Jacobian.
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