A question about a form of elements of G2 Let  $f$ be an automorphism of the octonions algebra.
Then $f(x)=x$ for $x\in \mathbb R$ and $f$ restricted to $Im \mathbb O$ is in $SO(7)$.
By the properties of the rotations there is an orthonormal basis: $e_1,e_2,...,e_7$ in $\mathbb O$ and real numbers $\phi_1,\phi_2,\phi_3$ such that:
$$
f(e_1)=e_1,\\
f(e_2)=\cos \phi_1  e_2-\sin \phi_1 e_3, \\
f(e_3)=\sin \phi_1  e_2+\cos \phi_1 e_3,\\
f(e_4)=\cos \phi_2 e_4-\sin \phi_2 e_5, \\
f(e_5)=\sin \phi_2 e_4+\cos \phi_2 e_5,\\
f(e_6)=\cos \phi_3 e_6-\sin \phi_3 e_7, \\
f(e_7)=\sin \phi_3 e_6+\cos \phi_3 e_7.
$$
Does there exist a Cayley's triple $(i,j,l)$ in $Im \mathbb  O$ (i.e. unit elements of $Im \mathbb O$  satisfying  $l\bot i,j,ij$) such that 
$$
e_1=i,\\
e_2=j,\\
e_3=k,\\
e_4=l,\\
e_5=il,\\
e_6=jl,\\
e_7=(ij)l.
$$
 A: You are essentially asking whether or not every element of $G_2$ is conjugate to an element in its maximal torus.  As you remark, every element of $G_2$ is conjugate to an element in the maximal torus of $\mathrm{SO}(7)$, that much is obvious from, as you say, the properties of rotations in dimension $7$.  
Now, for any compact, connected Lie group (which $G_2$ is), there is a general theorem that says that every element is conjugate to an element of a maximal torus in that group.  Since the maximal torus in $\mathrm{SO}(7)$ that you have written down contains a maximal torus in $G_2$, the answer to your question is 'yes'.  Indeed, you get a stronger result that you can get the above form where the angles $\phi_i$ are such that the above transformation belongs to $G_2$ (i.e., is an automorphism of $\mathbb{O}$.  I think that your basis is chosen so that this maximal torus is described by the relation
$$
\phi_1 +\phi_2 + \phi_3 = 0,
$$
but you should check this by checking the condition on these angles that makes the transformation an automorphism of $\mathbb{O}$.
