Stabilizer of $\operatorname{Aut}(B)$ acting on $0\to A\rightarrow B\rightarrow C\rightarrow 0$ I am reading an article on elementary homological algebra and have a trouble understanding one statement. Let $R$ be a ring and $A,B,C$ modules over $R$. Let $S$ be a set of exact sequences of the form
$$
0\rightarrow A\rightarrow B\rightarrow C \rightarrow 0
$$
The article says

$\operatorname{Aut}(B)$ acts on $S$ with stabilizer $1+\alpha \operatorname{Hom}(C,A)\beta$ where $\alpha,\beta$ are the maps fitting in the short exact sequence of the trivial extension
  $$
0 \rightarrow  A\stackrel{\alpha}{\rightarrow} A\oplus C \stackrel{\beta}{\rightarrow} C \rightarrow 0
$$

Firstly I don't quite understand what $1+\alpha \operatorname{Hom}(C,A)\beta$ means (what is $1+\dots$?) and secondly don't see why the stabilizer of $\operatorname{Aut}(B)$ is identified with the above set. 
Could anyone kindly explain what is going on? Thank you very much. 

Well, if $\gamma\in\hom(C,A)$, then $\alpha\gamma\beta:A\oplus C\to A\oplus C$: $(a,c)\mapsto (\gamma(c),0)$.
I think, in the highlighted part $B=A\oplus C$, then it makes sense, and $1$ means its identity, so that all $1+\alpha\gamma\beta$ will be an automorphism that fixes the exact sequence $(\alpha,\beta)$.
