= Solution
Covariance of the fundamental <gauge covariant derivative> requires $D_\mu^{f\prime}(g\chi)=gD_\mu^f\chi$ for every column $\chi$. Expanding the left side and cancelling $g\partial_\mu\chi$ gives $(\partial_\mu g)+A'_\mu g=gA_\mu$. Therefore the necessary and sufficient <Yang-Mills gauge transformation> is
$$
\boxed{A'_\mu=gA_\mu g^{-1}-(\partial_\mu g)g^{-1}.}
$$
In operator notation $D_\mu^{f\prime}=gD_\mu^fg^{-1}$. Expanding its <commutator> on an arbitrary column shows $[D_\mu^f,D_\nu^f]\chi=F_{\mu\nu}\chi$: the second derivatives cancel, as do the terms containing a single derivative of $\chi$. Conjugating this identity proves the <gauge field strength> transformation
$$
\boxed{F'_{\mu\nu}=gF_{\mu\nu}g^{-1}.}
$$
Thus the <gauge field strength> transforms homogeneously even though the potential has an inhomogeneous derivative term.
For the <adjoint scalar field>, write $B_\mu=(\partial_\mu g)g^{-1}$ and $\Phi'=g\Phi g^{-1}$. Differentiating the inverse matrix gives
$$
\partial_\mu\Phi'=g(\partial_\mu\Phi)g^{-1}+[B_\mu,\Phi'],\qquad [A'_\mu,\Phi']=g[A_\mu,\Phi]g^{-1}-[B_\mu,\Phi'].
$$
The extra terms cancel, proving \b[$D'_\mu\Phi'=g(D_\mu\Phi)g^{-1}$] for the <adjoint covariant derivative>. Its curvature is obtained by another direct expansion:
$$
\begin{aligned}
[D_\mu,D_\nu]\Phi
&=[\partial_\mu A_\nu-\partial_\nu A_\mu,\Phi]
+[A_\mu,[A_\nu,\Phi]]-[A_\nu,[A_\mu,\Phi]]\\
&=[\partial_\mu A_\nu-\partial_\nu A_\mu+[A_\mu,A_\nu],\Phi]
=\boxed{[F_{\mu\nu},\Phi]},
\end{aligned}
$$
where the <Jacobi identity> combines the last two terms.
For the equations of motion take compactly supported variations, and assume $e\ne0$, as required by the Lagrangian coefficient. Invariance of the <Killing form> implies
$$
\partial_\mu\kappa(X,Y)=\kappa(D_\mu X,Y)+\kappa(X,D_\mu Y).
$$
Indeed, the two connection terms sum to $\kappa([A_\mu,X],Y)+\kappa(X,[A_\mu,Y])=0$. This is the <gauge-covariant integration by parts> identity. The needed variations are
$$
\delta F_{\mu\nu}=D_\mu\delta A_\nu-D_\nu\delta A_\mu,\qquad
\delta(D_\mu\Phi)=D_\mu\delta\Phi+[\delta A_\mu,\Phi].
$$
Antisymmetry of the <gauge field strength> combines its two variation terms. Applying <gauge-covariant integration by parts> to the action $S=\int\mathcal L\,d^4x$ gives
$$
\begin{aligned}
\delta S=\int d^4x\biggl\{&-\frac1{e^2}\kappa(\delta A_\nu,D_\mu F^{\mu\nu})
-\kappa(\delta A_\nu,[\Phi,D^\nu\Phi])\\
&+\kappa(\delta\Phi,D_\mu D^\mu\Phi)\biggr\}.
\end{aligned}
$$
Here $\kappa([\delta A_\nu,\Phi],D^\nu\Phi)=\kappa(\delta A_\nu,[\Phi,D^\nu\Phi])$ fixes the sign of the scalar current. Since the <Killing form> is nondegenerate on a <semisimple Lie algebra>, independent variations give the coupled <Yang-Mills equations> and scalar equation
$$
\boxed{D_\mu F^{\mu\nu}+e^2[\Phi,D^\nu\Phi]=0,\qquad D_\mu D^\mu\Phi=0.}
$$
Equivalently, $D_\mu F^{\mu\nu}=e^2[D^\nu\Phi,\Phi]$. These signs use the stated <Minkowski metric>, with upper spatial derivatives $D^i=-D_i$.
It remains to verify every equation for the static <Bogomolny equations>. The hypotheses imply $F_{0i}=0$ and $D_0\Phi=0$, so the temporal gauge equation is automatically zero. The <gauge-theory Bianchi identity>
$$
D_iF_{jk}+D_jF_{ki}+D_kF_{ij}=0
$$
follows, for example, from the <Jacobi identity> for the fundamental <gauge covariant derivatives>, using their commutator $F_{ij}$. Define $B_i=\tfrac12\varepsilon_{ijk}F_{jk}$. Contracting the <gauge-theory Bianchi identity> with $\varepsilon_{ijk}$ gives $D_iB_i=0$. The <Bogomolny equations> say $B_i=eD_i\Phi$, so $e\ne0$ implies $D_iD_i\Phi=0$. In the static configuration $D_\mu D^\mu\Phi=-D_iD_i\Phi$, verifying the scalar equation.
For the spatial gauge equation, use $F^{ij}=F_{ij}$ and $D^j\Phi=-D_j\Phi$. The <Bogomolny equations>, the curvature commutator proved above, and the <contraction of two Levi-Civita symbols> give
$$
\begin{aligned}
D_iF_{ij}&=e\varepsilon_{ijk}D_iD_k\Phi
=\frac e2\varepsilon_{ijk}[D_i,D_k]\Phi
=\frac e2\varepsilon_{ijk}[F_{ik},\Phi]\\
&=\frac{e^2}2\varepsilon_{ijk}\varepsilon_{ik\ell}[D_\ell\Phi,\Phi]
=-e^2[D_j\Phi,\Phi]=e^2[\Phi,D_j\Phi],
\end{aligned}
$$
since $\varepsilon_{ijk}\varepsilon_{ik\ell}=-2\delta_{j\ell}$. This is exactly $D_iF^{ij}+e^2[\Phi,D^j\Phi]=0$. Thus \b[the static first-order Bogomolny system satisfies all four gauge equations and the scalar equation], with no extra ansatz or boundary assumption needed for this local implication. This is the mechanism by which <Bogomolny monopole equations imply Yang-Mills-Higgs equations>.
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