For the record, here's the complete solution. Let A,B,X be quaternions; A',B',X' are their conjugates. Here's some formulae we may need: S(A) = (A+A')/2 V(A) = (A-A')/2 A = S(A)+V(A) A' = S(A)-V(A) N(A) = AA'=A'A=|A|^2 N(V(A)) = N(A)-S(A)^2 V(A)^2 = -N(V(A)) A is a *unit* quaternion when N(A)=1. A is a 'pure' vector when S(A)=0. A is a 'pure unit vector' when N(A)=1 and S(A)=0. X=0 is the trivial solution of AX=XB. For X to have a nontrivial solution, N(AX)=N(A)N(X)=N(X)N(B)=N(XB) => N(A)=N(B). Also, since S(B) = S(1/X A X) = S(1/X (S(A)+V(A)) X) = S(1/X S(A) X + 1/X V(A) X) = S(S(A) X/X + 1/X V(A) X) = S(S(A) + 1/X V(A) X) = S(S(A)) = S(A) (Because A => (1/X V(A) X) merely rotates a pure vector into another pure vector w/o a scalar part.) So also for X to have a nontrivial solution, S(A)=S(B). Note also that N(A)=S(A)^2+N(V(A))=S(B)^2+N(V(B))=N(B), hence N(V(A))=N(V(B)). Let X = (N(V(A)) - V(A) V(B)) = |V(A)||V(B)| (1 - (V(A)/|V(A)|) (V(B)/|V(B)|)) (Because |V(A)|=|V(B)|, N(V(A))=|V(A)||V(B)|.) So this X is a real multiple of the X we found in the previous email when A,B were pure unit vectors. Indeed, V(A)/|V(A)|, V(B)/|V(B)| ARE pure unit vectors. Checking: AX = XB (S(A)+V(A)) X = X (S(B)+V(B)) S(A) X + V(A) X = S(B) X + X V(B) S(A) X + V(A) X = S(A) X + X V(B) V(A) X = X V(B) V(A) (N(V(A))-V(A)V(B)) = (N(V(A))-V(A)V(B)) V(B) V(A)N(V(A)) - V(A)^2 V(B) = N(V(A))V(B) - V(A)V(B)^2 N(V(A))V(A) - (-N(V(A))) V(B) = N(V(A)) V(B) - V(A) (-N(V(B))) N(V(A)) V(A) + N(V(A)) V(B) = N(V(A)) V(B) + N(V(B)) V(A) N(V(A)) (V(A)+V(B)) = N(V(A)) V(B) + N(V(A)) V(A) N(V(A)) (V(A)+V(B)) = N(V(A)) (V(B)+V(A)) V(A)+V(B) = V(B)+V(A) Check!