Q:

Is a "symmetric" definition of quaternions possible?

$\def\qi{\mathbf{i}}
\def\qj{\mathbf{j}}
\def\qk{\mathbf{k}}
\def\qr{\mathbf{r}}
$A quaternion $a+ b\qi + c\qj +d\qk$ and the defining equation $\qi^2 = \qj^2 = \qk^2 = \qi\qj\qk = -1$ are obviously visually non-symmetric in their treatment of the real part $a$ and the imaginary-like parts, $b$, $c$ and $d$. 
Is it possible to find a formulation that explicitly introduces the real part as a nominal unit vector, say $\qr$, such that the quaternion reads $a\qr+ b\qi + c\qj +d\qk$? And, more importantly, can we find a set of equations that are symmetric in these unit vectors such that the resulting structure is actually the same as the normal quaternions?
Put another way: is the asymmetry in the definition just "visual" because its convenient, or is the real part fundamentally different from the other three parts?

A:

Real and imaginary elements behave fundamentally different and it is a good thing this is represented in the defining relations of the standard basis $\{1,\mathbf{i},\mathbf{j},\mathbf{k}\}$ (as a real vector space). Asking for a more symmetric multiplication table here would be like asking for one instead of $\{1,i\}$ for the complex numbers $\mathbb{C}$. It is actually possible; if one uses $u=e^{\theta i}$ and $v=e^{-\theta i}$ instead for a basis, say with $\theta=\pi/4$, then the multiplication table is defined by 
$$u^2=-v, \quad v^2=-u, \quad uv=\frac{1}{\sqrt{2}}(u+v)$$ which is symmetric (i.e. swapping $u$ and $v$ in all the relations, they remain true) but unnecessarily so.
Reals square to positives while imaginary numbers square to negatives. This is also true for quaternions, and furthermore real numbers commute with everything while imaginary quaternions commute with reals and parallel vectors and anticommute with perpendicular vectors. This distinction is of fundamental importance when using quaternions to model 3D and 4D rotations, their most important application.
It is possible to create a basis for $\mathbb{H}$ which has a maximally symmetric multiplication table. However, this won't happen for a basis like $\{\mathbf{r},\mathbf{i},\mathbf{j},\mathbf{k}\}$ for two reasons: (1) it isn't even a basis, so you can't represent any real parts and imaginary parts of non-unique representations, and (2) $\{\mathbf{i},\mathbf{j},\mathbf{k}\}$ are an orthonormal basis of $\mathbb{R}^3$ whereas $\mathbf{r}$ is linearly dependent on them, so their multiplication table can't be symmetric!
The symmetry ought to be maximized when we use a basis $\{e^{\theta\mathbf{a}},e^{\theta\mathbf{b}},e^{\theta\mathbf{c}},e^{\theta\mathbf{d}}\}$ for a non-quadrantal angle $\theta$ and four vertices $\mathbf{a},\mathbf{b},\mathbf{c},\mathbf{d}$ of any regular tetrahedron in $\mathbb{R}^3$. If we again choose the angle $\theta=\pi/4$ (which is exactly between quadrantals and makes the nicest squaring relations, no other reason) we get a multiplication table consisting of relations like
$$ \left(e^{{\large \frac{\pi}{4}}\mathbf{a}}\right)^2=\frac{1}{2}(e^{{\large \frac{\pi}{4}}\mathbf{a}}-e^{{\large \frac{\pi}{4}}\mathbf{b}}-e^{{\large \frac{\pi}{4}}\mathbf{c}}-e^{{\large \frac{\pi}{4}}\mathbf{d}}) $$
$$ \left(e^{{\large \frac{\pi}{4}}\mathbf{a}}\right)\left(e^{{\large \frac{\pi}{4}}\mathbf{b}}\right)=\frac{2}{3}+\frac{1}{2}(\mathbf{a}+\mathbf{b})+\frac{3}{16}(\mathbf{c}-\mathbf{d}). $$
(In order to really be a multiplication table, I'd have to write $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$, $\mathbf{d}$ themselves in terms of $e^{{\large \frac{\pi}{4}}\mathbf{a}}$, $e^{{\large \frac{\pi}{4}}\mathbf{b}}$, $e^{{\large \frac{\pi}{4}}\mathbf{c}}$, $e^{{\large \frac{\pi}{4}}\mathbf{d}}$ which can be done simply but I just wanted to illustrate the symmetry.)
For the products of basis elements, the order of the differences like $\mathbf{c}-\mathbf{d}$ depends on how the four vertices are oriented in space. But other than this orientation-dependence, every permutation is possible, so the symmetry group of this multiplication table is $A_4$. (I suspect a symmetry group of $S_4$ would for commutativity but I don't immediately see why so maybe I'm wrong about that.)
On the other hand, to emphasize the difference between real and imaginary elements one ought to choose one real and three imaginary elements for a basis, in which case the greatest symmetry is when the three imaginary elements are orthonormal. Note, in fact, if $\{\mathbf{u},\mathbf{v},\mathbf{w}\}$ is any orthonormal basis for $\mathbb{R}^3$ (appropriately oriented), then $\{1,\mathbf{u},\mathbf{v},\mathbf{w}\}$ has the same multiplication table as $\{1,\mathbf{i},\mathbf{j},\mathbf{k}\}$ with a symmetry group of $A_3$ (which is $C_3$, so cyclic symmetry, hence the Brougham bridge relations).
The symmetry here comes from the symmetry of a regular tetrahedron within $\mathbb{R}^3$, not between real and imaginary elements, in fact this symmetry is hiding the fundamental asymmetry between real and imaginary elements.

