I have been trying to understand a sketch of a proof from P. Gabriel's article "Finite representation type is open".

Let $F$ be an algebraically closed field, and let $R$ be a finite dimensional $F$-algebra. Let $\operatorname{Mod}(R,F^n)$ denote the set of all $R$-module structures on $F^n$. Then $\operatorname{Mod}(R,F^n)$ has the structure of a scheme over $F$. Furthermore, the algebraic group $G = GL(F^n)$ acts on $\operatorname{Mod}(R,F^n)$, essentially by conjugation.

Let $M$ be a fixed $R$-module structure on $F^n$. The orbit $G.M$ is locally closed in $\operatorname{Mod}(R,F^n)$, and it is given its reduced scheme structure. Then the natural map $f\colon G \to G.M$, given by multiplication by $M$, is faithfully flat. Gabriel claims that $f$ is smooth. The way he draws this conclusion is by saying that the stabilizer of $M$ is $\operatorname{Aut}_R(M)$, which is an open subset of $\operatorname{Hom}_R(M,M)$, so it is smooth. My question has to do with understanding this.

I am assuming that the stabilizer $\operatorname{Stab}(M)$ is the fibre of $f$ over $M$, which has a natural scheme structure. Since $F$ is algebraically closed, and everything is finite type over $F$, to prove that $f$ is smooth, it suffices to show that this fibre is smooth over $F$. Note that this approach means that I have to take the natural scheme structure on the fibre.

The closed points of this fibre are definitely equivalent to the set of closed points of $\operatorname{Aut}_R(M)$. So, my questions are:

Why are $\operatorname{Stab}(M)$ and $\operatorname{Aut}_R(M)$ isomorphic as schemes?, and why is $\operatorname{Aut}_R(M)$ smooth over $F$? Any help would be appreciated.