= Solution
Requiring $D_\mu\phi$ to transform as $U^\dagger(D_\mu\phi)U$ and using the gauge-field transformation law gives
$$
D_\mu\phi=\partial_\mu\phi-ig[A_\mu,\phi].
$$
Indeed, differentiating $U^\dagger\phi U$ produces two inhomogeneous derivative terms, and those cancel against the inhomogeneous part of the transformed connection.
For $U=1+i\alpha^aT_a+O(\alpha^2)$,
$$
\phi'=U^\dagger\phi U
=\phi+i[\phi,\alpha^aT_a]+O(\alpha^2).
$$
If $[T_a,T_b]=if_{ab}{}^cT_c$, then
$$
\delta\phi^c=f_{ab}{}^c\alpha^a\phi^b.
$$
This is the infinitesimal <Adjoint representation of a Lie algebra>.
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