This seems like a very limited question, since if A^2 = -1, B^2 = -1 that means both A and B lie on the sphere S^2 = {a+bi+cj+dk ∊ H | a=0, b^2+c^2+d^2 = 1} of the quaternions H. And what is the general solution, in any case, of AX = XB ??? —Dan Henry Baker wrote: ----- I assume that this solution is known, but I've not seen it before -- e.g., in Wikipedia, etc. I am mystified why not. Find quaternion X s.t. AX=XB, A,B unit vector quaternions. A,B are unit vector quaternions => AA = BB = -1. Consider X=(1-AB). A(1-AB)=(1-AB)B A-AAB = B-ABB A-(-1)B = B-A(-1) A+B = B+A (Check!) -X = AB-1 is also a solution: A(AB-1)=(AB-1)B AAB-A = ABB-B -B-A = -A-B (Check!) Example: A = J, B = (cos(2t) + I sin(2t)) J, X = 1-AB = 2*cos(t)*(cos(t) - I sin(t)). Note that X is actually a square root, as is typical for quaternion rotations. If we want to normalize X, simply divide by 2*cos(t). If X isn't normalized, then denote X' as the conjugate of X, and compute B = X'AX/(X'X); since X'X=|X|^2, we never need to compute |1-AB|. -----