Quaternions: Linear Equation $Q_x = Q_1 \times Q_ x \times Q_2$ This question relates to Solving a transformation equation involving vectors and quaternions, however my equation was a bit different. After a bit of solving it arrives at ( I am trying to solve for $Q_x$ )
$$Q_{x}=Q_{known1}\times Q_{x}\times Q_{known2} $$
Now if we were not talking about quaternions, things would be simple. Here however I do not know how to go on and solve for $Q_x$
I have looked around a bit, but am still unsure on how to proceed, apart from maybe the (obvious but complicated?) way of spliting every quaternion to its elements and trying to derive a system of equations... Any help?
 A: I'm going to abbreviate the original equation to $ax=xb$ where all three symbols are quaternions. Instead of putting $a$ and $b$ on the same side, you might learn more by putting the $x$'s on the same side like this:
$$b=x^{-1}ax$$
That is, if $x$ is nonzero. We can already see that $0$ is always a solution to $ax=xb$ for any $a,b$. By normalizing $x$, we can also assume, without loss of generality, that $x$ is a unit quaternion (has norm $1$).
It's well-known that $\mathrm{Re}( x^{-1}qx)=\mathrm{Re}(q)$ so the above equation has no solutions if the real parts of $a$ and $b$ are unequal.
Furthermore, the subspace of quaternions with real part $0$ is invariant under conjugation. Rotation quaternions act transitively on the sets of quaternions with real part $0$ which share the same length. So we have a further constraint that $|\mathrm{Im}(a)|=|\mathrm{Im}(b)|$ for there to be any more nonzero solutions.
Under these conditions, you can find a rotation quaternion $x$ such that $x^{-1}\mathrm{Im}(a)x=\mathrm{Im}(b)$. Since $x^{-1}\mathrm{Re}(a)x=\mathrm{Re}(a)=\mathrm{Re}(b)$ already, we get that $x^{-1}ax=b$.
But this $x$ you found is not unique: if $u$ is any rotation around the axis determined by $b$'s complex part, $u^{-1}\mathrm{Im}(b)u=\mathrm{Im}(b)$, and there is such a $u$ for each angle of rotation you can think of. Then $(xu)^{-1}a(xu)=b$ for every such $u$.
So in summary, the solution set will be $0$ if the real parts of $a$ and $b$ don't match or if the pure quaternion parts of $a$ and $b$ have different length. Otherwise, you will also get one particular $x$ with the strategy I mentioned above, and further solutions will be $\{xu\mid u\in \mathrm{Stab}(\mathrm{Im}(b))\}$ where $\mathrm{Stab}(y)=\{q\in \mathbb H\mid q^{-1}yq=y\}$.
I think this is a complete set of solutions.
