This is something of an exercise is unwinding the definitions, but there's an interesting twist to it as well, so here's an outline of an answer:
Let $G$ be a connected $3$-dimensional Lie group (not necessarily unimodular) endowed with a left-invariant metric $q$. Let $\eta:TG\to\mathbb{R}^3$ be a left-invariant 1-form on $G$ such that $q = {^t}\eta\,\eta = {\eta_1}^2+{\eta_2}^2+{\eta_3}^2$ and fix an orientation on $G$ by requiring that $\eta_1\wedge\eta_2\wedge\eta_3$ be the oriented volume form. Regard the principal right $\mathrm{SO}(3)$-bundle $P_{SO}$ as the set of linear, oriented $q$-isometries $u:T_gG\to\mathbb{R}^3$, this is a trivial bundle isomorphic to $G\times\mathrm{SO}(3)$, where one identifies $(g,a)\in G\times\mathrm{SO}(3)$ with the element $u= a^{-1}\circ\eta_g:T_gG\to\mathbb{R}^3$ in $P_{SO}$. An automorphism of $P_{SO}=G\times\mathrm{SO}(3)$ is a smooth mapping of the form $\Phi(k,a)= \bigl(k,\ell(k)a\bigr)$, where $\ell:G\to\mathrm{SO}(3)$ is smooth, i.e., an element of the gauge group. It's useful to consider a slightly larger group, the extended automorphisms, which are mappings of the form $\Phi(k,a)= \bigl(gk,\ell(k)a\bigr)$ where $g\in G$ is fixed and $\ell:G\to\mathrm{SO}(3)$ is smooth.
A connection form $\omega$ on $P_{SO}$ is of the form
$$
\omega = a^{-1}\,\mathrm{d}a + a^{-1}\,\overline{\omega} \,a
$$
where $\overline{\omega} = - {^t}\overline{\omega}$ is a $1$-form on $G$ with values in $\mathfrak{so}(3)$. If $[\omega]$ belongs to $\mathscr{B}^G$, then, for each $g\in G$, there exists a smooth map $\ell_g:G\to\mathrm{SO}(3)$ so that
$$
L_g^*\overline{\omega} = \ell_g^{-1}\,\mathrm{d}\ell_g + \ell_g^{-1}\,\overline{\omega}\,\ell_g\,,\tag1
$$
where $L_g:G\to G$ is left-multiplication by $g\in G$.
Of course, this implies that
$$
L_g^*\bigl(\mathrm{d}\overline\omega + \overline\omega\wedge\overline\omega\bigr)
= \ell_g^{-1}\,\bigl(\mathrm{d}\overline\omega + \overline\omega\wedge\overline\omega\bigr)\,\ell_g\tag2
$$
For $x\in\mathbb{R}^3$, let $\langle x\rangle\in\mathfrak{so}(3)$ satisfy $\langle x\rangle y = x\times y$. Note the useful identity $a^{-1}\langle x\rangle a = \langle a^{-1}x\rangle$ for $a\in\mathrm{SO}(3)$. It follows that there is a smooth mapping $F:G\to M_{3\times3}(\mathbb{R})$ such that
$$
\mathrm{d}\overline\omega + \overline\omega\wedge\overline\omega
= \bigl\langle F\,{\ast}\eta\bigr\rangle\qquad\text{where}\quad
{\ast}\eta = \begin{pmatrix}\eta_2\wedge\eta_3\\\eta_3\wedge\eta_1\\\eta_1\wedge\eta_2\end{pmatrix}.
$$
Since $\eta$ is left-invariant, we have $L_g^*({\ast}\eta) = {\ast}\eta$, so equation (2) becomes
$$
L_g^*F = \ell_g^{-1} F.\tag3
$$
Of course, this implies $L_g^*\bigl({^t}FF\bigr) = {^t}FF$, i.e., the symmetric matrix ${^t}FF$ is constant. Consequently, $(\det F)^2 = \det\bigl({^t}\!FF\bigr)$ is constant. Since $G$ is connected, $\det F$ is constant. Note that the rank of $F$ is also constant. It follows that there is a matrix $R\in\mathrm{SO}(3)$ such that ${^t}\!(FR)(FR)$ is a constant diagonal matrix. Replacing $\eta$ by $R^{-1}\eta$, we can reduce to the case that ${^t}\!FF$ itself is diagonal, with nonnegative and non-increasing entries down the diagonal.
The special cases in which $F$ has rank $0$ or $1$ proceeds by a separate argument (see the remark below), so suppose $F$ has rank at least $2$. Then $F$ can be written in the form $F = QD$ where $D$ is a constant diagonal matrix with at least 2 positive entries and $Q:G\to SO(3)$ is smooth. Equation (3) then implies that $L_g^*Q = \ell_g^{-1}\,Q$, i.e., $\ell_g = Q(L_g^*Q)^{-1}$.
In particular $\ell_g:G\to\mathrm{SO}(3)$ is unique and smooth in $g$.
Thus, for any $g\in G$, equation (1) can be re-arranged to become
$$
L_g^*\bigl(Q^{-1}\,\mathrm{d}Q + Q^{-1}\overline\omega Q\bigr)
= Q^{-1}\,\mathrm{d}Q + Q^{-1}\overline\omega Q,
$$
i.e., the $1$-form $\hat\omega = Q^{-1}\,\mathrm{d}Q + Q^{-1}\overline\omega Q$ on $G$ is left-invariant. Consequently, the connection form
$$
\tilde\omega = a^{-1}\,\mathrm{d}a + a^{-1}\hat\omega a
$$
on $P_{SO} = G\times\mathrm{SO}(3)$ (which is gauge equivalent to $\omega$) is invariant under $L_g\times \mathrm{id}_{\mathrm{SO}(3)}$.
Thus, setting $\Phi(k,a) = (k,Qa)$, we see that $\Phi^*\omega = \tilde\omega$, so setting
$$\phi(g) = \Phi\circ(L_g\times \mathrm{id}_{\mathrm{SO}(3)})\circ\Phi^{-1},
$$
we have that $\phi(g)^*\omega = \omega$, and $\phi$ is a homomorphism of $G$ into the extended automorphism group of $P_{SO}$.
N.B. When $F$ has rank $0$ or $1$, there is no well-defined $Q$, so there's not a well-defined $\hat\omega$. Moreover, $\ell_g$ is not uniquely defined by (1), so a different argument is needed. (This is the case in which the connection $\omega$ itself has nontrivial automorphisms.) The analysis in these cases is left to the reader.
