Let $X$ be a projective smooth curve over an algebraically closed field $\mathbb k$ of any characteristic. Let also $G$ be a reductive group over $\mathbb k$. (Probably) following Ramanathan, a $G$-bundle $\mathcal E$ on $X$ is called semistable if for all parabolic subgroups $P$ of $G$, all reductions $\mathcal E_{P}$ of $\mathcal E$ to $P$ and all dominant characters $\chi:P\rightarrow \mathbb G_m$, one has $$ \deg(\mathcal E_P\times_P \mathbb A^1_{\chi})\le 0. $$

(to be clear $\mathcal E_P\times_P \mathbb A^1_{\chi}$ here denotes the natural linear bundle on $X$ obtained from $\mathcal E_P$ by applying $\chi$ to the structure group $P$).

Now, this definition is not superintuitive for me, but looking at the case $G=GL_n$ I convinced myself that it's pretty natural (and agrees with the slopes' one).

**So**, my question is: is it true that the group of $\mathbb k$-points $\mathrm{Aut}(\mathcal E)(\mathbb k)$ of the automorphisms of $\mathcal E$ always obtains an emdedding in $G(\mathbb k)$ if $\mathcal E$ is semistable? Or if not, is there any sort of description of the possible automorphism groups of semistable $E$'s? Can they be bounded by some other group? (In a sense that there is a group $A_G$ over $\mathbb k$ such that any automorphism group admits an embedding to $A_G$, and $A_G$ does not depend, or depends in an easy way, on $X$)

For example in the case of $GL_n$, since stable bundles are simple, homs between two stable bundles are either $\mathbb k$ or 0, and having a polystable bundle $\mathcal E=\bigoplus_i \mathcal S_i^{\oplus k_i}$ the automorphisms are given by $\prod_{i}GL_{k_i}$, which is naturally a subgroup of $GL_n$. I'm not completely sure what happens for more complicated semistable bundles.