= Solution
For a gauge-field configuration, spatial symmetry means <symmetry up to gauge transformation>: the pullback of the fields by a spatial isometry lies in their original <gauge orbit>. In the anti-Hermitian convention,
$$
A^g=gAg^{-1}-dg\,g^{-1},\qquad
\Phi^g=g\Phi g^{-1}.
$$
The potential and charged fields need not be strictly fixed in a particular gauge; their gauge-invariant observables must be fixed.
For example, the Abelian potential $A=Bx\,dy$ describes a uniform magnetic field. Translating $x$ by $a$ adds $Ba\,dy=d(Bay)$, a gauge change, although the potential itself changes. For an axially symmetric vortex $\Phi=f(r)e^{iN\theta}$, $A=a(r)d\theta$, rotation through $\beta$ multiplies $\Phi$ by $e^{iN\beta}$; a constant $U(1)$ transformation compensates. These illustrate why literal invariance of the representative is too strong.
For the <'t Hooft-Polyakov monopole>, a spatial rotation $R\in SO(3)$ has an $SU(2)$ lift $g_R$ with adjoint action $R$. The two lifts differ by a central sign and act identically on adjoint fields. The <hedgehog ansatz for a monopole> satisfies
$$
\Phi^a(Rx)=R_{ab}\Phi^b(x),\qquad
(R^*A)_i^a(x)=R_{ji}A_j^a(Rx)=R_{ab}A_i^b(x).
$$
The second identity follows from invariance of $\epsilon_{iaj}$ under simultaneous rotations of all three indices. Thus the pulled-back fields are obtained by the constant gauge transformation $g_R$; applying $g_R^{-1}$ restores the representative. In particular $|\Phi|^2$ and the energy density are radial even though $\Phi^a=f(r)x^a/r$ is not literally invariant under rotating space alone.
For the field equations, use the invariant inner product with $[e_a,e_b]=\epsilon_{abc}e_c$. Since $B_i^aB_i^a=\tfrac12F_{ij}^aF_{ij}^a$, a variation $a_j=\delta A_j$, $\psi=\delta\Phi$ gives
$$
\delta V_\lambda
=2\int_{\mathbb R^3}
\left\langle-D_iF_{ij}+[\Phi,D_j\Phi],a_j\right\rangle
-2\left\langle D_iD_i\Phi+\lambda\Phi,\psi\right\rangle\,d^3x.
$$
Here the scalar kinetic contribution to the gauge variation uses
$\langle D_j\Phi,[a_j,\Phi]\rangle
=\langle[\Phi,D_j\Phi],a_j\rangle$. The <Euler-Lagrange field equations> for the energy actually printed are
$$
\boxed{D_iF_{ij}=[\Phi,D_j\Phi],\qquad
D_iD_i\Phi+\lambda\Phi=0.}
$$
The PDF has $\lambda(1-|\Phi|^2)$, without a square. This is not the usual nonnegative monopole potential with vacuum $|\Phi|=1$. If the usual $\lambda(1-|\Phi|^2)^2$ were intended, only the scalar equation would change to
$$
D_iD_i\Phi+2\lambda(1-|\Phi|^2)\Phi=0.
$$
The final requested proof at $\lambda=0$ is identical for either interpretation; we do not silently alter the printed energy.
To derive the covariant derivative of the radial fields, first differentiate the Higgs profile:
$$
\partial_k\Phi^a
=\frac fr\delta_{ak}+\frac1r\left(\frac fr\right)'x^ax^k.
$$
The commutator term is
$$
[A_k,\Phi]^a
=\epsilon_{abc}\epsilon_{kbj}x^jx^c\,\frac{\alpha f}{r}
=\frac{\alpha f}{r}(r^2\delta_{ak}-x^ax^k).
$$
Adding the two terms gives exactly
$$
\boxed{D_k\Phi^a=
\left(\frac fr+r\alpha f\right)\delta_{ak}
+\frac1r\left[\left(\frac fr\right)'-\alpha f\right]x^ax^k.}
$$
The radial eigenvalue of the supplied magnetic tensor is $2\alpha+r^2\alpha^2$, while that of this covariant derivative is $f'$. Its transverse coefficients are respectively $r\alpha'+2\alpha$ and $f/r+r\alpha f$. Consequently $B=-D\Phi$ reduces to
$$
\boxed{f'=-2\alpha-r^2\alpha^2,\qquad
r\alpha'+2\alpha=-\frac fr-r\alpha f.}
$$
Equivalently, the <hedgehog monopole equations in radial profile variables> with $K=1+r^2\alpha$ are
$$
\boxed{K'=-fK,\qquad f'=\frac{1-K^2}{r^2}.}
$$
These are two equations rather than an overdetermined system: radial and transverse projections span the two invariant tensor structures. For a regular monopole in this hedgehog sector, smoothness at the origin gives $f(0)=0$, $K(0)=1$, while the normalized vacuum conditions give $f\to1$, $K\to0$ at infinity; solving the ODEs is not needed here.
Finally, prove that the <Bogomolny monopole equations imply Yang-Mills-Higgs equations>, without restricting to the radial ansatz. The three-dimensional <gauge-theory Bianchi identity> is $D_iB_i=0$. If $B_i=-D_i\Phi$, then
$$
D_iD_i\Phi=-D_iB_i=0,
$$
which is the scalar equation at $\lambda=0$. For the gauge equation,
$$
\begin{aligned}
D_iF_{ij}
&=-\epsilon_{ijk}D_iD_k\Phi\\
&=-\frac12\epsilon_{ijk}[F_{ik},\Phi]\\
&=[B_j,\Phi]=[\Phi,D_j\Phi].
\end{aligned}
$$
In the second line antisymmetry eliminates the symmetric derivative part; the curvature commutator is $[D_i,D_k]\Phi=[F_{ik},\Phi]$. The next equality uses
$\epsilon_{ijk}\epsilon_{ik\ell}=-2\delta_{j\ell}$ and $F_{ik}=\epsilon_{ik\ell}B_\ell$. These are exactly both field equations at $\lambda=0$, so \b[every smooth Bogomolny solution satisfies them], not merely those of the hedgehog form.
For finite-energy monopoles, the associated energy argument is also transparent:
$$
V_0=\int|B+D\Phi|^2\,d^3x
-2\int_{S^2_\infty}\langle\Phi,B_i\rangle\,dS_i.
$$
The <gauge-theory Bianchi identity> converts $\langle B_i,D_i\Phi\rangle$ into this boundary term. At fixed asymptotic magnetic charge, $B=-D\Phi$ saturates the <Bogomolny bound>. This complements the local differential proof above and explains the energy-minimizing role of the first-order equations.
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