Automorphisms of the exceptional Jordan algebra preserve the determinant and trace.

Question

I am trying to show that the automorphism of the exceptional Jordan algebra $\mathbb{J}_3(\mathbb{O})$ preserve the determinant and trace. This algebra consists of $3x3$ Hermitian matrices over octonions: $$ X = \begin{pmatrix} a & z & \overline{y} \\ \overline{z} & b & x \\ y & \overline{x} & c \end{pmatrix} $$ with product defined by $X \circ Y = \frac12 (XY+YX)$. What I want to understand is, why does any automorphism of $\mathbb{J}_3(\mathbb{O})$ preserve $\det(X)$ and $\operatorname{tr}(X)$. I was thinking towards the Cayley-Hamilton theorem (namely, its form for the $3\times 3$ matrices): $$ X^3 - \operatorname{tr}(X)X^2 - \frac12 (\operatorname{tr}(X^2) - \operatorname{tr}(X)^2)X - \det(X)I = 0, $$ but I don't know what to do next. Can you please suggest anything?