 Solution

For the convention in the question, a finite Yang-Mills gauge transformation acts covariantly on the field strength:or with $U$ and $U^{†}$ exchanged if the opposite convention is used for $D_{\mu }$. Cyclicity of the matrix trace givesso 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.