Just to elaborate on what is already in the comments, the algebra automorphisms of $\mathbb H$ act transitively on the set of pairs $(u,v)$ where $u$ and $v$ are imaginary quaternions of unit length that are orthogonal to one another.
To see this, I will include here some remarks on $\mathbb H$ and its automorphisms. Part of the OP's concern seems to be that it is not a priori automatic that metric concepts in $\mathbb H$ such as unit length or orthogonality (and hence the notion of being imaginary, since the imaginary quaternions are the orthogonal complement to $\mathbb R$ in $\mathbb H$)
are preserved by Aut$(\mathbb H)$, and so one of my goals is to show that this concern is not necessary. Indeed, this geometry is intrinsic to the quaternions, as we will see.
(This is not coincidence: Hamilton was led to his discovery by trying to algebraize the geometry of $\mathbb R^3$.)
Note first that imaginary quaternions are characterized by the condition that
$\overline{u} = - u$, and thus for such quaternions, $|u|^2 = -u^2$. Thus if $u$ is imaginary, $u^2$ is a non-positive real number. Converesly,
if $u^2$ is a negative real number,
then one sees that $u$ is imaginary (exercise), and so the imaginary quaternions are also
characterized by having non-positive real squares. In particular, the set of imaginary
quaternions is preserved by Aut$(\mathbb H)$.
On imaginary quaternions, the inner product $u\overline{v} + v\overline{u}$ is simply
$u v + u v$, and so is also preserved by Aut$(\mathbb H)$. In particular, metric concepts like "length one" and "orthogonal" are preserved by Aut$(\mathbb H)$.
If $u$ and $v$ are unit length orthogonal imaginary quaternions, we then have that
$u^2 = v^2 = -1$ (unit length condition) and that $u v = - v u$ (orthogonality condition).
Thus, from the defining relations of $\mathbb H$, we obtain an algebra map
$\mathbb H \to \mathbb H$ that maps $i$ to $u$ and $j$ to $v$ (and then $k$ to $u v$).
This map is non-zero (since $u$ and $v$ are non-zero, having unit length), and hence
is necessarily injective ($\mathbb H$ is a division ring, hence has no non-trivial ideals),
and thus in fact bijective (source and target are of the same dimension).
An automorphism of $\mathbb H$ is determined by its values on $i$ and $j$ (since they generate
$\mathbb H$), and so the previous discussion shows that in fact Aut$(\mathbb H)$ is the
same as the group or permutations of pairs $(u,v)$ of orthogonal pairs of unit
vectors in the imaginary quaternions (also known as $\mathbb R^3$).
This group is well-known: it is precisely $SO(3)$. (If you like, $u$ and $v$ determine
uniquely a mutually orthogonal vector --- their quaternionic product $u v$ --- which can be characterized geometrically in terms of $u$ and $v$ via the right hand rule; thus pairs $(u,v)$ are the same as positively oriented orthonormal bases of $\mathbb R^3$, permutations of which are precisely the group $SO(3)$.)
Incidentally, it is not coincidence that Aut$(\mathbb H) = SO(3)$.
Namely, there is a natural map $\mathbb H^{\times} \to $ Aut$(\mathbb H)$ (where
$\mathbb H^{\times}$ means the non-zero --- equivalently invertible --- quaternions),
given by mapping $q$ to the automorphism $x \mapsto q x q^{-1}$.
The kernel of this map is precisely the centre, and so it induces an injection
$\mathbb H^{\times}/\mathbb R^{\times} \hookrightarrow $ Aut$(\mathbb H)$.
Now the source of this map can be identified with the quotient of the unit quaternions
(which form a copy of $SU(2)$) by $\pm 1$, and of course $SU(2)/\{\pm 1\} =
SO(3)$. On the other hand, this injection is in fact a bijection (i.e. any automorphism
of $\mathbb H$ is inner), by the Skolem--Noether theorem. This puts the description of Aut$(\mathbb H)$ obtained above into a more general perspective.
The answer to both questions is yes. Just like Alford, Granville and Pomerance's proof that there are infinitely many Carmichael numbers, my proof that there are infinitely many Carmichael numbers where the minimal polynomial splits completely is essentially a constructive proof.
Both proofs need to jump through some hoops to get provability that you would not need in practice.
Grantham, Jon. There are infinitely many Perrin pseudoprimes. J. Number Theory 130 (2010), no. 5, 1117–1128.
Best Answer
You should probably read "On Quaternions and Octonions" by J.H. Conway and D.A. Smith
P.S. It's "octonion" not "octonian"
Edit: The first thing you will find is a discussion of integral numbers. For the complex numbers you have $\mathbb{Z}[i]$ (aka Gaussian integers) which is an $A_1\times A_1$ lattice. You also have $\mathbb{Z}[\omega]$ (aka Eisenstein integers) which is an $A_2$-lattice. For quaternions you have the integral quaternions (as above) which is an $A_1^4$ lattice and you also have the Hurwitz integral quaternions (adjoin $(1+i+j+k)/2$) which as a lattice is $D_4$. The Hurwitz numbers have "division with small remainder" property which makes them better. For the octonions you might start with doubling the Hurwitz integers to get a $D_4\times D_4$ lattice. Then you might add more and get $D_8$. However the octavian integers are an $E_8$ lattice. They are not unique (there are seven versions). These have several good properties:
Every left or right ideal is principal.
Every ideal is two-sided.
Then there is a discussion of prime factorisations.
Finally the automorphism group of the octaves has a simple subgroup of index 2 and order 12096. This group is $G_2(2)$.