Does the $3+4$ coloring $\chi$ determine the homomorphism $\rho_{\chi}$?

Consider the seed:

$$ \mathcal S = (\mathcal I, \Gamma, \Pi) $$

where $\mathcal I$ is a geometric realization, $\Gamma \subset \mathcal I$ is the distinguished $1$-skeleton, and $\Pi$ is the piecewise isometry group generated by "Rubik style moves" on the instrument $\mathscr I= (\mathcal I, \Gamma).$

In the undecorated setting I learned that there is no natural nontrivial homomorphism from the abstract $2 \times 2$ Rubik's cube group to $\mathrm{PGL}(3,2)$. I realized that the correct source of projective symmetry is not the bare group, rather a decorated realization of $\mathcal S$, which is what I was implicitly assuming, and should have made clear in that post.

A Fano decorated seed is a tuple:

$$ \mathcal S_\chi=(\mathcal I,\Gamma,\Pi,\chi), $$

where \(\chi\) is a \(3+4\) partition of seven distinguished cells of \((\mathcal I,\Gamma)\), written $$ \mathcal C=\lbrace m_x,m_y,m_z,s_{xy},s_{xz},s_{yz},s_{xyz} \rbrace. $$

Here $m_x,m_y,m_z$ denote the three meridian cells, and $s_{xy},s_{xz},s_{yz},s_{xyz}$ denote the remaining four cells.

We identify these seven cells with the seven nonzero vectors of $\mathbf F_2^3$ by the rule $$ m_x=e_1,\qquad m_y=e_2,\qquad m_z=e_3, $$ $$ s_{xy}=e_1+e_2,\qquad s_{xz}=e_1+e_3,\qquad s_{yz}=e_2+e_3,\qquad s_{xyz}=e_1+e_2+e_3. $$

This determines a canonical Fano plane incidence structure on \(\mathcal C\).

The line set \(\mathcal L\) associated to \(\mathcal C\) is the collection of seven triples $$ \mathcal L= \Big\lbrace \lbrace m_x,m_y,s_{xy}\rbrace, \lbrace m_x,m_z,s_{xz} \rbrace, \lbrace m_y,m_z,s_{yz} \rbrace, \lbrace s_{xy},s_{xz},s_{yz} \rbrace, $$ $$ \lbrace m_x,s_{yz},s_{xyz} \rbrace, \lbrace m_y,s_{xz},s_{xyz} \rbrace, \lbrace m_z,s_{xy},s_{xyz} \rbrace \Big \rbrace. $$ Equivalently, under the identification \(\mathcal C\cong \mathbf F_2^3\setminus\{0\}\), the lines are exactly the zero-sum triples $$ {a,b,a+b}, $$ or equivalently the triples \(\{a,b,c\}\) satisfying $$ a+b+c=0. $$

Thus $(\mathcal C,\mathcal L)$ is canonically isomorphic to the Fano plane. There is a canonical homomorphism $$ \rho_\chi:\Pi\longrightarrow \mathrm{PGL}(3,2). $$

It is important to state that the projective symmetry is a property of the decorated realization, not of the undecorated group in isolation. One could easily define a different color partition $\chi$ allowing for the manifestation of a different symmetry group than the Fano plane symmetry.

I created this depiction of the Fano decorated seed $ \mathcal S_\chi=(\mathcal I,\Gamma,\Pi,\chi) $:

Question. Does the $3+4$ coloring $\chi$ determine the homomorphism $$ \rho_\chi:\Pi\longrightarrow \mathrm{PGL}(3,2) $$ intrinsically, uniquely up to conjugacy, as the induced incidence-preserving action of $\Pi$ on the associated Fano plane?