The first nontrivial irreducible representation of $G\_2$ is of 7-dimensional, and the first nontrivial representation of $F\_4$ is of 26-dimensional.


My question is: how much is known about the nilpotent orbits in these representations? Any classification? or the answer is very easy, there are only nilpotent orbits, one is zero and another is nonzero orbit.



Nilpotent orbits in representations of exceptional groups
As per the OP's comment, we are to assume that $\mathrm{G}\_2$ and $\mathrm{F}\_4$ mean the complex simple Lie groups.


Let's start with $\mathrm{G}\_2\subset\mathrm{SO}(7,\mathbb{C})$, in its standard representation on $\mathbb{C}^7$, which is the vector space $V= \mathrm{Im}(\mathbb{O}^\mathbb{C})\subset \mathbb{O}^\mathbb{C}$, where $\mathbb{O}^\mathbb{C}$ is the (non-associative) algebra of octonions over the ground field $\mathbb{C}$. Let $\mathbf{1}\in\mathbb{O}^\mathbb{C}$ be the multiplicative unit, and let $\langle,\rangle$ be the non-degenerate inner product on $\mathbb{O}^\mathbb{C}$ that satisfies $\langle xy, xy\rangle = \langle x, x\rangle \langle y, y\rangle$ for all $x,y\in\mathbb{O}^\mathbb{C}$. Then $V = \mathbf{1}^\perp$, and, for $x\in V$, we have $x^2 = -\langle x,x\rangle\,\mathbf{1}$, and $\mathrm{G}\_2$ is the group of automorphisms of $\mathbb{O}^\mathbb{C}$. It is well-known that $\mathrm{G}\_2$ acts simply transitively on the orthonormal triples $(x\_1,x\_2,x\_3)$ in $V$ that satisfy $\langle x\_1x\_2,x\_3\rangle = 0$.


Since $\mathrm{G}\_2$ preserves the non-degenerate quadratic form $\langle,\rangle$ on $\mathbb{C}^7$, each orbit of $\mathrm{G}\_2$ lies in a level set of the non-degenerate quadratic form.


First, each level set $\langle x,x\rangle = \lambda \not=0$ is a single $\mathrm{G}\_2$ orbit, as this follows from the fact that $\langle x,x\rangle = 1$ is a single $\mathrm{G}\_2$-orbit, which, in turn follows from the above well-known characterization of $\mathrm{G}\_2$.


Second, $\mathrm{G}\_2$ fixes $0\in V$, so it only remains to understand the orbit structure on the level set $\langle x,x\rangle =0$ minus the origin. Suppose that $x\not=0$ satisfies $\langle x,x\rangle =0$. Then $x$ lies in a non-degenerate $2$-plane $E\subset V$ with an orthonormal basis $(x\_1,x\_2)$. Select an element $x\_3\in V$ of unit norm and perpendicular to $x\_1$, $x\_2$, and $x\_1x\_2$, and let $F$ be the span of $x\_1$, $x\_2, x\_3$. From the above well-known result, $\mathrm{G}\_2$ contains a subgroup $H$ that preserves $F$ and acts as $\mathrm{O}(3,\mathbb{C})$ on $F$. Since $\mathrm{O}(3,\mathbb{C})$ acts transitively on the nonzero null elements in $\mathbb{C}^3$, it follows that $H$ acts transitively on the nonzero null vectors in $F$.
In particular, we can assume, after an action by an element of $H$, that $x = x\_1 + i x\_2$, and so, again, by the well-known result, we see that $\mathrm{G}\_2$ must act transitively on the level set $\langle x,x\rangle =0$ minus the origin.


Hence $\mathrm{G}\_2$ has exactly one non-closed orbit, i.e., the level set $\langle x,x\rangle =0$ minus the origin, and its closure is the level set $\langle x,x\rangle =0$.


The $\mathrm{F}\_4$ case is a little more involved, and there are more non-closed orbits in its $26$-dimensional representation, but it can be understood in a similar way by identifying its representation of dimension $26$ as $\mathbb{J}\_0$, the elements of the complex, $27$-dimensional exceptional Jordan algebra $\mathbb{J}$ that have zero trace, by using the structure of the Jordan algebra. The key point is that $\mathrm{F}\_4$ preserves both a quadratic and a cubic form on $\mathbb{J}\_0$ and the non-closed orbits all lie the simultaneous zero set of those two forms.


