= Solution
The <permutation> inverse is $\rho_{g^{-1}}$: applying the two maps successively sends $\omega$ to $(\omega g)g^{-1}=\omega$. With the paper's right-action convention, <permutation> products also act on the right, and
$$
\omega(\rho_g\rho_h)=(\omega g)h=\omega(gh)=\omega\rho_{gh}.
$$
Thus $\boxed{\rho(gh)=\rho(g)\rho(h)}$. If function composition is written on the left instead, the same right action is represented by $g\mapsto\rho_{g^{-1}}$; keeping these conventions consistent avoids an artificial reversal. The kernel consists precisely of elements fixing every point, so
$$
\boxed{\ker\rho=\bigcap_{\omega\in\Omega}G_\omega.}
$$
The rotational tetrahedral <group> acts transitively on its four vertices. A <rotation> fixing a vertex must preserve its opposite face and rotate its three vertices cyclically; its <stabilizer> has order three. The <orbit-stabilizer theorem> gives $|G|=4\cdot3=12$. The rotations are the identity, two nontrivial rotations about each of the four vertex-to-opposite-face axes, and a half-turn about each of the three opposite-edge-midpoint axes. Therefore the counts by element order are
$$
\boxed{\begin{array}{c|rrrrrr}n&1&2&3&4&6&12\\\#&1&3&8&0&0&0\end{array}.}
$$
A <rotation> fixing all vertices fixes three <linearly independent> vertex vectors and is the identity. The vertex <permutation> action is therefore faithful. The listed rotations induce the identity, 3-cycles and double transpositions, all even permutations. Its injective image has order 12, the full order of $A_4$. Hence $\boxed{G\cong A_4}$.
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