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 $$ \mathrm{d}\overline\omega + \overline\omega\wedge\overline\omega = \bigl\langle F\,{\ast}\eta\bigr\rangle\qquad\text{where}\quad {\ast}\eta = \begin{pmatrix}\eta_2\wedge\eta_3\\\eta_3\wedge\eta_1\\\eta_1\wedge\eta_2\end{pmatrix}. $$ Because $\eta$ is left-invariant, $L_g^*({\ast}\eta) = {\ast}\eta$, so the above 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 given $g\in G$, 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 $L_g\times \mathrm{id}_{\mathrm{SO}(3)}$.

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.

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.