This is not an answer but a long comment on the Robert Bryant's answer. I that that $H=Im(j)$.

One has to show that $(jA)[q,\xi q]=(jA)[1,\xi q]$ for any $\xi,q\in \mathbb{O}$ with $|q|=1$. Using that $A_{11},A_{22}\in \mathbb{R}$ and $Re((ab)c)=Re(a(bc))$ for any octnonions $a,b,c$ one has \begin{eqnarray} (jA)[q,\xi q]=\\ \bar qA_{11}q+\overline{(\xi q)}A_{22}(\xi q)+2Re(\bar qA_{12}(\xi q))=\\ A_{11}+A_{22}|\xi|^2+2Re(((\xi q)\bar q) A_{12})=\\ A_{11}+A_{22}|\xi|^2+2Re(\xi(q\bar q) A_{12})=\\ A_{11}+A_{22}|\xi|^2+2Re( A_{12}\xi)=\\ (jA)[1,\xi]. \end{eqnarray} QED.

This is not an answer but a long comment on the Robert Bryant's answer. I think that $H=Im(j)$.

One has to show that $(jA)[q,\xi q]=(jA)[1,\xi q]$ for any $\xi,q\in \mathbb{O}$ with $|q|=1$. Using that $A_{11},A_{22}\in \mathbb{R}$ and $Re((ab)c)=Re(a(bc))$ for any octnonions $a,b,c$ one has \begin{eqnarray} (jA)[q,\xi q]=\\ \bar qA_{11}q+\overline{(\xi q)}A_{22}(\xi q)+2Re(\bar qA_{12}(\xi q))=\\ A_{11}+A_{22}|\xi|^2+2Re(((\xi q)\bar q) A_{12})=\\ A_{11}+A_{22}|\xi|^2+2Re(\xi(q\bar q) A_{12})=\\ A_{11}+A_{22}|\xi|^2+2Re( A_{12}\xi)=\\ (jA)[1,\xi]. \end{eqnarray} QED.