I work over the complexe numbers, but the story is similar over any fields (provided you choose the split octonions). Let us identify the Albert algebra $A$ with the algebra of self-adjoint $3*3$ matrices with octonionic coefficients. The hyperplane given by $\mathrm{Tr}(X) = 0$ is stabilized by $F_4 \times \mathbb{G}_m$ and the action of $F_4 \times \mathbb{G}_m$ on this hyperplane can be devided into two types.

First : the orbits located in the discriminant hypersurface (that is the set of matrices of rank less or equal to $2$). They are:

_the zero of $A$,

_the set of matrice of rank $1$ (denote it by $Z_0$, it has dimension $16$),

_an orbit which closure contains the previous one and that has dimension $17$,

_the (open part) of a certain restricted tangent variety to $Z_0$,

_the discriminant hypersurface minus the previous one and the closure of the third one : it has dimension $25$.

Note that the equation of the discriminant hypersurface is $6\mathrm{det}(X)^2-9(\mathrm{Tr}(X^2))^3 = 0$.

Now, let's turn to the orbits located outside of the discriminant hypersurface. There is a one dimensional family of them: they are the hypersurfaces given by the equation $6t\mathrm{det}(X)^2-s(\mathrm{Tr}(X^2))^3 = 0$, for $[s,t] \in \mathbb{P}^1 \backslash [1,9]$.

Al of this is discussed at lentgh and proved in details in proposition 3.5 of https://arxiv.org/pdf/math/0306328.pdf

I work over the complexe numbers, but the story is similar over any fields (provided you choose the split octonions). Let us identify the Albert algebra $A$ with the algebra of self-adjoint $3*3$ matrices with octonionic coefficients. The hyperplane given by $\mathrm{Tr}(X) = 0$ is stabilized by $F_4 \times \mathbb{G}_m$ and the action of $F_4 \times \mathbb{G}_m$ on this hyperplane can be devided into two types.

First : the orbits located in the discriminant hypersurface (that is the set of matrices of rank less or equal to $2$). They are:

_the zero of $A$,

_the set of matrice of rank $1$ (denote it by $Z_0$, it has dimension $16$),

_an orbit which closure contains the previous one and that has dimension $17$,

_the (open part) of a certain restricted tangent variety to $Z_0$,

_the discriminant hypersurface minus the previous one and the closure of the third one : it has dimension $25$.

Note that the equation of the discriminant hypersurface is $6\mathrm{det}(X)^2-9(\mathrm{Tr}(X^2))^3 = 0$.

Now, let's turn to the orbits located outside of the discriminant hypersurface. There is a one dimensional family of them: they are the hypersurfaces given by the equation $6t\mathrm{det}(X)^2-s(\mathrm{Tr}(X^2))^3 = 0$, for $[s,t] \in \mathbb{P}^1 \backslash [1,9]$.

All of this is discussed at lentgh and proved in details in proposition 3.5 of https://arxiv.org/pdf/math/0306328.pdf

I work over the complexe numbers, but the story is similar over any fields (provided you choose the split octonions). Let us identify the Albert algebra $A$ with the algebra of self-adjoint $3*3$ matrices with octonionic coefficients. The hyperplane given by $\mathrm{Tr}(X) = 0$ is stabilized by $F_4 \times \mathbb{G}_m$ and the action of $F_4 \times \mathbb{G}_m$ on this hyperplane has $4$ orbits:

i) the zero of $A$,

ii) the set of matrices of "rank 1", which has dimension $16$,

iii) the set of matrices of "rank 2", which has dimension $25$,

iv) the set of matrices of "rank 3", which has dimension $26$.

The stabilizers are known, but I don't remember them off-hand. Interesting references are provided by the work of Landsberg and Manivel on the subject. See for instance : https://arxiv.org/pdf/math/9810140.pdf, https://www-fourier.ujf-grenoble.fr/sites/default/files/ref_477.pdf

I work over the complexe numbers, but the story is similar over any fields (provided you choose the split octonions). Let us identify the Albert algebra $A$ with the algebra of self-adjoint $3*3$ matrices with octonionic coefficients. The hyperplane given by $\mathrm{Tr}(X) = 0$ is stabilized by $F_4 \times \mathbb{G}_m$ and the action of $F_4 \times \mathbb{G}_m$ on this hyperplane can be devided into two types.

First : the orbits located in the discriminant hypersurface (that is the set of matrices of rank less or equal to $2$). They are:

_the zero of $A$,

_the set of matrice of rank $1$ (denote it by $Z_0$, it has dimension $16$),

_an orbit which closure contains the previous one and that has dimension $17$,

_the (open part) of a certain restricted tangent variety to $Z_0$,

_the discriminant hypersurface minus the previous one and the closure of the third one : it has dimension $25$.

Note that the equation of the discriminant hypersurface is $6\mathrm{det}(X)^2-9(\mathrm{Tr}(X^2))^3 = 0$.

Now, let's turn to the orbits located outside of the discriminant hypersurface. There is a one dimensional family of them: they are the hypersurfaces given by the equation $6t\mathrm{det}(X)^2-s(\mathrm{Tr}(X^2))^3 = 0$, for $[s,t] \in \mathbb{P}^1 \backslash [1,9]$.

Al of this is discussed at lentgh and proved in details in proposition 3.5 of https://arxiv.org/pdf/math/0306328.pdf