The geometric interpretation of quaternion multiplication is fundamentally 4-dimensional (unlike quaternion conjugation, which can be considered as an action on $\Bbb{R}^3$).
Let's start with an easy case. Say $q=a+bi$ with $b \neq 0$, $a^2+b^2=1$. That is, $q$ is a non-real unit quaternion in the subalgebra of $\Bbb{H}$ generated by $i$. What affect does multiplying by $q$ have on an arbitrary quaternion $r$?
First of all, if $r$ also lies in the subalgebra generated by $i$, then we can consider multiplication of $q$ and $r$ to be ordinary complex multiplication; that is, multiplication by $q$ rotates $r$ by $\theta$ in the $\{1, i\}$ plane.
Secondly, if $r$ lies in the orthogonal complement of that subalgebra, $r=cj+dk$, we can write $r=(c+di)j=j(c-di)$. The first of these representations can be used to left-multiply by $q$ via ordinary complex multiplication; the second one can be used to right-multiply. In either case, multiplying by $q$ rotates $r$ by $\theta$ in the $\{j, k\}$-plane; however, the sign difference means that the two multiplications rotate in opposite directions from each other.
We can then find the effect of multiplying $q$ by an arbitrary quaternion by projecting that quaternion into these two planes. That is, an arbitrary quaternion will have its $\{1, i\}$-projection and $\{j, k\}$-projection both rotated by $\theta$ when it is multiplied by $q$; the direction of the $\{j, k\}$-rotation, but not of the $\{1, i\}$-rotation, will be affected by whether we're left-multiplying or right-multiplying by $q$.
The general case works similarly. For any unreal unit quaternion $q$ that makes an angle $\theta$ with the real axis, multiplication by $q$ rotates by $\theta$ in the $\{1, q\}$-plane, and also rotates by $\theta$ in its orthogonal complement. The direction of the first rotation is fixed, but the direction of the second rotation depends on whether we're multiplying by $q$ on the left or on the right. You can see this just by noticing that any unreal quaternion generates a 2-dimensional subalgebra isomorphic to $\Bbb{C}$, making the previous few paragraphs work in general after some relabeling.
This also gives you a way to see why quaternion conjugation works the way it does on $\Bbb{R}^3$. If $q$ makes an angle $\theta$ with the real axis, then the map $r \mapsto qrq^{-1}$:
- is the identity map in the $\{1, q\}$-plane, since that plane forms a commutative subalgebra of $\Bbb{H}$
- involves a rotation by $2\theta$ in its orthogonal complement. Both $q$ and $q^{-1}$ rotate quaternions in $\{1, q\}^{\perp}$ by $\theta$. If they were multiplied on the same side, those rotations would have to cancel out; since left-multiplication behaves oppositely to right-multiplication, this means they must reinforce each other. Since the orthogonal complement of $\{1, q\}$ is orthogonal to $1$, it is pure imaginary, so we've reproduced the fact that quaternion conjugation corresponds to a double-angle rotation in $\Bbb{R}^3$ (when identified with $\Im(\Bbb{H})$).
Note also that multiplying by a general quaternion involves scaling by the norm of that quaternion, but conjugation conveniently causes the norms of $q$ and $q^{-1}$ to cancel.
Given the axes of the geometric body (hereafter: "body axes"), described with respect to some fixed global axes, you should be able to derive the corresponding rotation matrix that would rotate the global axes onto the body axes and vice versa.
In other words, if the body axes are $b_1, b_2, b_3$ and the global axes are $g_1, g_2, g_3$, then you want a rotation $R$ that does
$$R(g_1) = b_1, R(g_2) = b_2, R(g_3) = b_3$$
Or perhaps vice versa. Since inversion is the same as transposition for rotations, vice versa isn't hard to do either.
Now, again, let the body axes be described in terms of the global axes like so:
$$b_1 = B^{11} g_1 + B^{21} g_2 + B^{31} g_3$$
And similarly for $b_2, b_3$. Then, the matrix is basically computed for you. If you wrote out the above relations in matrix language, using the $g$ basis, you would have
$$R \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} B^{11} \\ B^{21} \\ B^{31}\end{bmatrix}$$
You should recognize now that this is just the first column of the matrix $R$. The columns of $R$ are just the body axes' unit vectors, expressed in terms of the global basis vectors.
Now that the matrix is known, you can use well-established algorithms to translate that into quaternions, if you should need to.
Best Answer
There is a definitely a connection, but how good of a connection depends on your expectations. There is a lot of sexy interplay between the metric geometry using Clifford algebras and the quaternions. Clifford (or "geometric") algebras encode geometric information about the underlying space.
I'd like to mention three such algebras. First of all, $C\ell_{0,2}(\mathbb{R})=\mathbb{H}$.
Secondly, for ordinary Euclidean $\mathbb{R}^3$, the algebra $C\ell_{3,0}(\mathbb{R})$ contains $\mathbb{H}$ as a subalgebra of elements which enact rotations in the traditional quaternion way. (Of course, LOTS of Clifford algebras contain copies $\mathbb{H}$ but this is the one that does it in the "natural way".) These are the so-called rotors in the algebra.
Lastly, moving up to Minkowski space with $C\ell_{1,3}(\mathbb{R})$, the rotors are a bit different than the quaternions. The encompass both spatial rotations and Lorentz boosts and mixtures of the two. I think the phenomenon you observed might be a quaternion-like shadow of these rotors.
In your post it looked like you were reading a paper using biquaternions (complex quaternions), and I can recommend another good paper on that if you haven't found it already: check out Lambek's If Hamilton Had Prevailed: Quaternions in Physics (1995) Math.Intelligencer. He is generally known for good exposition and he's knowledgable about this topic in particular (it was part of his dissertation).
P.S.: If you're wondering about $C\ell_{3,1}(\mathbb{R})$, it's also interesting, but it is nonisomorphic to $C\ell_{1,3}(\mathbb{R})$ as $\mathbb{R}$ algebras. For the purposes of physics, I don't think any evidence has arisen to choose one as "better".