Let $G<\mathrm{GL_n}$ be a simple linear algebraic group defined over a finite field $K$. Let $\mathfrak{g}$ be its Lie algebra. Assume $\mathfrak{g}$ is simple.

Is it necessarily the case that there is no subspace $\mathfrak{v}\subset \mathfrak{g}$ with $0<\dim(\mathfrak{v})<\dim(\mathfrak{g})$ such that $\mathfrak{v}$ is invariant under $\mathrm{Ad}_g$ for every $g\in G(K)$?

Note: it is clear that there is no $\mathfrak{v}\subset \mathfrak{g}$ with $0<\dim(\mathfrak{v})<\dim(\mathfrak{g})$ such that $\mathfrak{v}$ is invariant under $\mathrm{Ad}$ for every $g\in G(\overline{K})$. It is also clear that the answer to the question above is "yes" when the number of elements of $K$ is larger than a constant depending only on $n$: since $\mathfrak{v}$ is not an ideal, there is a $v\in\mathfrak{v}$ such that all $g\in G(\overline{K})$ such that $\mathrm{Ad}_g(v)\in \mathfrak{v}$ lie in a proper subvariety of $G$.

Note 2: A friend has just proposed over the breakfast table that there are linear algebraic groups with no non-trivial rational points over $K$. That would obviously imply an answer of "no" to my question. I am not convinced that such a thing is really possible, at least not when we are talking about the group $G(K)$, $G$ simple (as opposed to more exotic groups of Lie type). EDIT: as per Venkataramana's comment below, this situation cannot, in fact, occur for $G$ simple.