= Solution
Place the cube vertices at $(\pm1,\pm1,\pm1)$. The four vertices whose coordinate product is $+1$ form $T$; those with product $-1$ form $T'$. Within either <set>, two distinct vertices differ in exactly two coordinates, so every edge has length $2\sqrt2$ and each <set> is a <regular tetrahedron>.
A member of the <cube rotation group> is a <signed permutation matrix> of <determinant> one. There are $3!$ permutations of the coordinate axes and four choices of signs for each permutation that give <determinant> one, so $|G|=24$. Such a <matrix> maps a coordinate-product class to a coordinate-product class. Moreover, the <rotation> through a quarter-turn $(x,y,z)\mapsto(-y,x,z)$ interchanges the classes. Thus the <orbit> of $T$ consists of exactly $T,T'$. By the <orbit-stabilizer theorem>, the <tetrahedron stabilizer in the cube rotation group> has order
$$
\boxed{|G_T|=24/2=12.}
$$
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