= Solution
Count the pairs $(g,\omega)$ for which $g$ fixes $\omega$. Counting first by <group elements> gives $\sum_g\operatorname{Fix}(g)$, while counting first by points gives $\sum_\omega|\operatorname{Stab}(\omega)|$. On a <group orbit> of size $r$, <orbit-stabilizer theorem> makes each <stabilizer> have size $|G|/r$, so that <group orbit> contributes $|G|$ to the second sum. Dividing by $|G|$ proves <Burnside's lemma>:
$$
\boxed{\text{number of orbits}=\frac1{|G|}\sum_{g\in G}\operatorname{Fix}(g).}
$$
A <tetrahedron> has four vertices. A <rotational symmetry> is determined by the <image> of one vertex, with four possibilities, and the cyclic orientation of the three vertices around it, with three possibilities. All these choices are realizable, so its <rotational symmetry group of a tetrahedron> has order $12$. More explicitly, its elements are the identity, the two nontrivial <rotational symmetries> around each of four vertex-to-opposite-face axes, and the half-turn about each of three axes through opposite edge midpoints. Thus there is \b[one element of order 1, eight of order 3, and three of order 2]. The induced <permutations> of vertices are the <even permutations>, identifying the <group> with $A_4$.
Apply the <group action> to the <set> of <edge colourings>. A colouring is fixed by a <rotational symmetry> precisely when its colour is constant on each cycle of the induced edge <permutation>. The identity has six edge cycles and fixes $4^6$ colourings. A vertex-axis <rotational symmetry> has two cycles of length three, fixing $4^2$. An opposite-edge half-turn fixes the two edges on its axis as <sets> and exchanges the other four in two pairs, giving four cycles and $4^4$ fixed colourings. Therefore
$$
\boxed{\frac{4^6+8\cdot4^2+3\cdot4^4}{12}=416.}
$$
Reflections are excluded because the equivalence in the source uses <rotational symmetries>.
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