Torsors over discrete group schemes $G$ are finite Galois cover of $G$ Let $X$ be a connected scheme over a field $k$ and $G$ be a finite discrete group scheme over $k$.
By $G$-torsor $Y$ over $X$ I mean there is a fpcq cover $U$ of $X$ and $Y$ after base change over $U$ becomes isomorphic to the trivial $G$-torsor over $U$.
Let $T \rightarrow X$ be $G$-torsor over $X$. It is well-known that $T$ is then a finite etale Galois cover of $G$ but I've never seen a proof of this fact.
My question is why is $\operatorname{Aut}(T/X)$ is isomorphic to $G$ in this case? Also why is the cover Galois?
 A: First, we note that trivial $G$-torsors for a finite group scheme $G$ are automatically finite etale covers: a trivial $G$-torsor over a connected scheme $Z$ over a field $k$ is just $Z\times_k G\to Z$, which is the base change of $G\to \operatorname{Spec}k$ by the map $Z\to \operatorname{Spec} k$. As $G\to \operatorname{Spec} k$ is etale (it's smooth of relative dimension zero), $Z\times_k G\to Z$ is also etale because being etale is preserved under base change.
Now for the case of a general $G$-torsor $T\to X$. Let $U\to X$ be a fpqc covering so that $U\times_X T \to U$ is a trivial $G$-torsor. By the previous paragraph, we have that this is finite etale. The properties of being etale and being finite are both preserved by fpqc descent (etale: EGA IV4, 17.7.3(ii), finite: EGA IV2, 2.7.1(xv), thanks to Bjorn Poonen for a very handy reference table on page 302 here), so $T\to X$ is also finite etale. It remains to check the Galois condition.
Here I think you're missing a few assumptions compared to what I know of this subject. For me, a torsor comes with a $G$-action, and a Galois covering should be connected. Once you know that there's a $G$-action on $T\to X$, you can check that $Aut_X(T)$ is actually $G$ fairly quickly: $Aut_X(T)\hookrightarrow Aut_U(U\times_X T)$ as any automorphism of $T\to X$ gives an automorphism of $U\times_X T\to U$, and no nonidentity automorphism of $T\to X$ can induce the identity on $U\times_X T \to U$ since fpqc maps are surjective. I don't see ways to cook up this $G$-action from just the conditions in your post, nor any way to guarantee that $Y$ is connected (just take $Y=X\times G$).
