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 question about a form of elements of G2
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 $\mathbb{O}$.


