We have an affine transformation of the cube which leaves the faces as congruent rhombuses. As before we have that in plain coordinates the basis vectors are
(a, b, b), (b, a, b), (b, b, a) where a² + 2b² = 1 (since their lengths are 1) but the oblique dihedral angle between two faces is 120°. Using the Gibbs cross product we require that the vectors perpendicular to two sides are at an angle of 120°.
(a, b, b)×(b, a, b) = (bb−ab, bb−ab, aa-bb)
(b, a, b)×(b, b, a) = (aa−bb, bb−ab, bb−ab)
The angle between the two normals is the external dihedral angle between the two faces which is 180° − 120° = 60°. Two vectors p and q are at 60° iff
p∙q = ½ (length p)(length q) iff (p∙q) = ½ (p∙p) (since (p∙p) = (q∙q) in this case).
These equations are homogeneous so we shed some weight and let b=−1 first.
(a, −1, −1)×(−1, a, −1) = (1+a, 1+a, aa-1)
(−1, a, −1)×(−1, −1, a) = (aa−1, 1+1, 1+a)
p∙q = (a³ + a² − a − 1) + (2a + 1) + (a³ + a² − a − 1)
= 2a³ + 2a² − 1
p∙p = q∙q = 2a² + 4a + 2 + a⁴ -2a² + 1
= a⁴ + 4a + 1

(p∙q) = ½ (p∙p) ↔ 2a³ + 2a² − 1 = ½(a⁴ + 4a + 1) ↔ a⁴ − 4a³ − 4a² + 4a + 3 = 0
(Schemishly (define (p a) (+ 3 (* a (+ 4 (* a (+ -4 (* a (+ -4 a)))))))))

Root 1: -1
Root 2: -0.64575131106459
Root 3: 1
Root 4: 4.64575131106459

With that as a crib we notice that our polynomial is (a² − 1)(a² − 4a − 3) and that the root of interest is
(4 + √(16 + 12))/2 = 2 + √7.

Returning to the homogeneous version we have a = (2 + √7)b and a² + 2b² = 1.
((2 + √7)b)² + 2b² = (11 + 4√7 + 2)b² = 1 whence
b = 1/√(13 + 4√7) = 0.20592089929727178 and a = (2 + √7)/√(13 + 4√7) = 0.9566572878859