In a simple Euclidean Jordan algebra, the Jordan-automorphism group of
the algebra is the subgroup of the cone of squares' automorphism group
that fixes the Jordan unit element. This is written down, for example,
in Theorem 8 of Gowda, but the result goes all the way back to
Vinberg, even for non-simple EJAs.
For $\mathbb{R}$ and $\mathbb{C}$, that's enough to squeeze the result
out of Theorem 2 in Schneider. He gives you the operators that
preserve the cones of squares (the PSD cones), namely,
$$
\operatorname{Aut}\left(\mathcal{H}^{n}_{+}\left(\mathbb{R}\right)\right)
=
\left\lbrace X \mapsto U^{T}XU \ \middle|\ U \in \mathbb{GL}_{n}\left(\mathbb{R}\right) \right\rbrace
$$
and
$$
\begin{aligned}
\operatorname{Aut}\left(\mathcal{H}^{n}_{+}\left(\mathbb{C}\right)\right)
&=
\left\lbrace X \mapsto U^{*}XU \ \middle|\ U \in \mathbb{GL}_{n}\left(\mathbb{C}\right) \right\rbrace\\
&\cup
\left\lbrace X \mapsto U^{*}\overline{X}U \ \middle|\ U \in \mathbb{GL}_{n}\left(\mathbb{C}\right) \right\rbrace.
\end{aligned}
$$
Imposing the additional condition that these preserve the identity matrix,
$$
\begin{aligned}
\operatorname{JAut}\left(\mathcal{H}^{n}\left(\mathbb{R}\right)\right)
&=
\left\lbrace X \mapsto U^{T}XU \ \middle|\ U \in \mathbb{R}^{n \times n}, U^{T}U = I \right\rbrace,\\
\operatorname{JAut}\left(\mathcal{H}^{n}\left(\mathbb{C}\right)\right)
&=
\left\lbrace X \mapsto U^{*}XU \ \middle|\ U \in \mathbb{C}^{n \times n}, U^{*}U = I \right\rbrace\\
&\cup
\left\lbrace X \mapsto U^{*}\overline{X}U \ \middle|\ U \in \mathbb{C}^{n \times n}, U^{*}U = I \right\rbrace.
\end{aligned}
$$
An analogous result based on the same sort of inertia theorem holds
over the quaternions, $\mathbb{H}$. The book by Rodman contains
enough of the spectral theory to show that,
$$
\operatorname{Aut}\left(\mathcal{H}^{n}_{+}\left(\mathbb{H}\right)\right)
=
\left\lbrace X \mapsto U^{*}XU \ \middle|\ U \in \mathbb{GL}_{n}\left(\mathbb{H}\right) \right\rbrace,
$$
from which it follows that
$$
\operatorname{JAut}\left(\mathcal{H}^{n}\left(\mathbb{H}\right)\right)
=
\left\lbrace X \mapsto U^{*}XU \ \middle|\ U \in \mathbb{H}^{n \times n}, U^{*}U = I \right\rbrace.
$$
I've just posted a preprint where I work out all of these
details. This method has the advantage that you wind up knowing the
cone automorphisms as well, but there is a more direct approach that
can be used to verify the results. Theorem 6.5 in Huang characterizes the matrix Jordan-automorphism groups over $\mathbb{R}$, $\mathbb{C}$, and
$\mathbb{H}$, assuming that you know the linear automorphism groups of
$\mathbb{R}$, $\mathbb{C}$, and $\mathbb{H}$ as algebras over
$\mathbb{R}$. Here again Rodman comes to the rescue with the
prerequisite result for $\mathbb{H}$. In any case, we eventually
deduce the same Jordan-automorphism groups that we do using the
stabilizer subgroup approach.
