This isn't comprehensive, this is just some thought I had about quaternion commutativity.

$q = q^u e_u$, for $e_u \in \{ e_0, e_1, e_2, e_3 \}$
also depicted $e_u \in \{ 1, i, j, k \}$
also as $q = (w, v) = (w, x, y, z)$, where $v \in \mathbb{R}^3$, $w \in \mathbb{R}$.

$e_u$ holds the following product rules:
$ij=k=-ji, jk=i=-kj, ki=j=-ik$
$i^2 = j^2 = k^2 = -1$

Notice I will be using vector inner product $a \cdot b = a_x b_x + a_y b_y + a_z b_z$ and cross product $a \times b = (a_y b_z - a_z b_y, a_z b_x - a_x b_z, a_x b_y - a_y b_x)$

addition: $q_1 + q_2 = (w_1 + w_2, v_1 + v_2)$
scaling by reals: $s q = (s w, s v)$
negation: $-q = (-w, -v)$

multiplication:
$q_1 q_2 = (w_1 + x_1 i + y_1 j + z_1 k) \cdot (w_2 + x_2 i + y_2 j + z_2 k)$
$= w_1 w_2 - x_1 x_2 - y_1 y_2 - z_1 z_2 + (x_1 w_2 + w_1 x_2 + y_1 z_2 - z_1 y_2) i + (y_1 w_2 + w_1 y_2 + z_1 x_2 - x_1 z_2) j + (z_1 w_2 + w_1 z_2 + x_1 y_2 - y_1 x_2) k $

multiplication, in vector form:
$q_1 q_2 = (w_1 w_2 - v_1 \cdot v_2, w_1 v_2 + w_2 v_1 + v_1 \times v_2)$

conjugation: $\bar{q} = (w, -v)$
inner product: $q_1 \cdot q_2 = q_1 \bar{q}_2 = w_1 w_2 + v_1 \cdot v_2$
norm: $|q|^2 = q \cdot q = w^2 + x^2 + y^2 + z^2$
inverse: $q^{-1} = \bar{q} / |q|^2$

Commutation:
$[q_1, q_2] = q_1 q_2 - q_2 q_1$
$= (w_1 w_2 - v_1 \cdot v_2, w_1 v_2 + w_2 v_1 + v_1 \times v_2) - (w_2 w_1 - v_2 \cdot v_1, w_2 v_1 + w_1 v_2 + v_2 \times v_1)$
$= ( w_1 w_2 - v_1 \cdot v_2 - w_2 w_1 + v_2 \cdot v_1, , w_1 v_2 + w_2 v_1 + v_1 \times v_2 - w_2 v_1 - w_1 v_2 - v_2 \times v_1 )$
$= (0, 2 v_1 \times v_2)$

Because the commutation is not zero, we know that $q_1 q_2 \ne q_2 q_1$

But if you want to convert a rotation into the frame of another rotation, you can use the trick $q'_2 = (q_1)^{-1} q_2 (q_1)$ such that: $q_1 ((q_1)^{-1} q_2 (q_1)) = q_2 q_1$

exponent: $exp(q) = exp(w) (cos|v|, \frac{v}{|v|} sin|v|)$
logarithm: $log(q) = (log|q|, \frac{v}{|v|} acos(\frac{w}{|q|}))$

Alright so how well does commutativity line up with the exponential map?
Let's assume $q = (cos(\theta), sin(\theta) n)$, for $\theta \ge 0$ and $|n| = 1$
From this we see $|q| = 1$

$log(q_1) + log(q_2) = (log|q_1|, \frac{v_1}{|v_1|} acos(\frac{w_1}{|q_1|})) + (log|q_2|, \frac{v_2}{|v_2|} acos(\frac{w_2}{|q_2|})) $
$= ( 0, n_1 acos(cos(\theta_1)) + n_2 acos(cos(\theta_2)) ) $
$= (0, n_1 \theta_1 + n_2 \theta_2)$

Disclaimer that this only holds true for $\theta$ in the domain of $acos$.

$log(q_1 q_2) = log( (w_1 w_2 - v_1 \cdot v_2, w_1 v_2 + w_2 v_1 + v_1 \times v_2) )$
$= log( ( cos(\theta_1 ) cos(\theta_2) - sin(\theta_1) sin(\theta_2) n_1 \cdot n_2, cos(\theta_1) sin(\theta_2) n_2 + cos(\theta_2) sin(\theta_1) n_1 + sin(\theta_1) sin(\theta_2) n_1 \times n_2 ) )$
$= ( 0, ( cos(\theta_1) sin(\theta_2) n_2 + cos(\theta_2) sin(\theta_1) n_1 + sin(\theta_1) sin(\theta_2) n_1 \times n_2 ) acos ( cos(\theta_1) cos(\theta_2) - sin(\theta_1) sin(\theta_2) n_1 \cdot n_2 ) )$

Maybe the other way?
$exp((0,\theta_1 n_1) + (0, \theta_2 n_2))$ vs $exp((0, \theta_1 n_1)) exp((0, \theta_2 n_2))$

exponent: $exp(q) = exp(w) (cos|v|, \frac{v}{|v|} sin|v|)$
$exp((0, \theta_1 n_1) + (0, \theta_2 n_2))$
$= ( cos|\theta_1 n_1 + \theta_2 n_2|, \frac{\theta_1 n_1 + \theta_2 n_2}{|\theta_1 n_1 + \theta_2 n_2|} sin|\theta_1 n_1 + \theta_2 n_2| )$

$exp((0, \theta_1 n_1)) exp((0, \theta_2 n_2))$
$= (cos|\theta_1 n_1|, \frac{n_1}{|n_1|} sin|\theta_1 n_1|) (cos|\theta_2 n_2|, \frac{n_2}{|n_2|} sin|\theta_2 n_2|) $
$= (cos(\theta_1), n_1 sin(\theta_1)) (cos(\theta_2), n_2 sin(\theta_2)) $
$= ( cos(\theta_1) cos(\theta_2) - sin(\theta_1) sin(\theta_2) n_1 \cdot n_2, cos(\theta_1) sin(\theta_2) n_2 + cos(\theta_2) sin(\theta_1) n_1 + sin(\theta_1) sin(\theta_2) n_1 \times n_2 )$