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 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](http://en.wikipedia.org/wiki/Skolem-Noether_theorem).  This puts the description of Aut$(\mathbb H^3)$ obtained above into a more general perspective.