Discovery 2019-01-01
Let $u$ be order 3 octonion of norm $1$ and trace $-1$ (in case of characteristic 2 we have $-1=1$; norm is $x\bar x$, trace is $x+\bar x$). Then element $L_uR_{u^{-1}}$ is automorphism. This way we obtain order 3 generators of $G_2(q)$. I tested it in GAP for $q=2,3,4,5,7,8,9$. I don't know how to prove it yet but it seems to be rather easy. I define finite octonions as $\mathbb H_q+\mathbb H_q\iota$ using Cayley-Dickson formula. We define quaternions $\mathbb H_q=M_2\mathbb F_q$ as matrices $2\times 2$ over finite field. Conjugation is defined as adjugate matrix - exchange elements on the diagonal and reverse sign off-diagonal. The complex numbers algebra is finite field $\mathbb F_{q^2}$ represented by subalgebra generated by matrix $\pmatrix{n&1\\ 1&0}$ of order $q+1$ when $n$ is order $q-1$ generator of the field multiplicative group. This matrix satisfy equation $x^2=nx+1$.
Having $\mathbb C_q=\mathbb F_{q^2}$ we can also define quaternions using Cayley-Dickson formula $\mathbb H_q=\mathbb C_q+\mathbb C_qj$. Conjugation is then defined as $\overline{x+yj}=\bar{x}-yj$. It coincides with Frobenius automorphism $x\to x^q$ of the field $\mathbb F_{q^2}$.
Similar law is valid for compact octonions. Take $u=e^{2πi/3}$ for some imaginary octonion $i$. Then $L_uR_{u^{-1}}$ is automorphism fixing complex plane $\langle1,i\rangle$ and rotating by angle $\frac{2π}{3}$ in perpendicular space.
End of discovery