Let $Q=(Q\_0,Q\_1)$ be a quiver and $q\in Q\_0$ a chosen vertex. Let $d$ be a dimension vector with $d\_q=1$ and let $\theta\in \mathbb R^{Q\_0}$ be a $d$-generic stability parameter. Let $M$ be a $\theta$-stable $Q$-representation of dimension $d$. Let $f$ be an automorphism of $M$ which is the identity on the component $M\_q$. Is it true that $f$ must necessarily be the identity of $M$?



Are there non-trivial automorphisms of stable framed quiver representations?
All automorphisms of the quiver representation you mention are by multiplication by scalars. So as $f$ is the identity on $M\_q$ it must be the identity on $M$.


Under your assumptions, $g=f-\mathrm{Id}\_M$ is an endomorphism of $M$ with nonzero kernel. The morphism $g\colon M\rightarrow \mathrm{Im}(g)$ is surjective, and so if $\mathrm{Im}(g)\neq 0$ the slope of this representation is at least as great as that of $M$, by stability. On the other hand, $\mathrm{Im}(g)$ is a submodule of $M$, so its slope is no greater than that of $M$. Putting these two inequalities together we get that $\mathrm{Im}(g)$ is either zero, or has the same slope as $M$.


Under the genericity assumptions, $M$ cannot have a subrepresentation of the same slope, since the dimension vector of such a subrepresentation would have to divide $d$, and $d\_q=1$. So $\mathrm{Im}(g)=0$ and $f=\mathrm{Id}\_M$.


