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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 $$ e_1=i,\\ e_2=j,\\ e_3=k,\\ e_4=l,\\ e_5=il,\\ e_6=jl,\\ e_7=(ij)l. $$

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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 $\mathbb{O}$.

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  • $\begingroup$ It seems that $\phi_1+\phi_2=\phi_3$ in order $f(j)f(l)=f(jl)$. $\endgroup$
    – emiliocba
    Feb 2, 2023 at 10:15
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    $\begingroup$ @emiliocba: I think you are correct about the sign of $\phi_3$. I didn't actually check it when I wrote the answer back then. The OP's 'standard basis' is different from mine, that's why I suggested that the OP check. I like my basis better. (I would use $e_7 = i(jl)$ instead of $e_7=(ij)l$ because then left multiplication by $e_1=i$ induces the 'correct' complex structure on $e_1^\perp\simeq \mathbb{R}^6 = \mathbb{C}^3$.) $\endgroup$ Feb 2, 2023 at 12:18
  • $\begingroup$ Nice! By the way, Do you know any reference that deals with explicit maximal tori of G_2 realized as the automorphism of octonions? I still do not understand how I can pick the three pairs of elements in a fixed standard basis of octonions $\endgroup$
    – emiliocba
    Feb 2, 2023 at 16:23

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