Rigidity of extensions Let $\newcommand\mc{\mathcal}\mc E : 0 \to A \to E \to B \to 0$ be an extension of abelian groups (or more generally take any abelian category $\mc A$), this defines an equivalence class of extensions $[\mc E] \in \DeclareMathOperator{\Ext}{Ext} \Ext^1(B,A)$.
Define $\DeclareMathOperator{\Aut}{Aut}\Aut([\mc E])$ to be the automorphisms of this equivalence class, i.e. if $\mc E_i : 0 \to A \to E_i \to B \to 0$ for $i=1,2$ are two extensions in the equivalence class $[\mc E]$
and $\varphi : E_1 \to E_2$ defines an isomorphism of extensions.
Is there a relation between maps $\DeclareMathOperator{\Mor}{Mor}\Mor(B,A)$ and automorphisms $\Aut([\mc E])$?
For example, can one say that if $\Mor(B,A) = 0$ then $\Aut([\mc E]) = 0$, so there is a unique isomorphism between two extensions in the equivalence class $[\mc E]$ (Rigidity)?
My attempt was to look at the difference $\varphi_1 - \varphi_2$ of two maps in $\Aut([\mc E])$ and to construct a map $B \to A$ in a similar way to the snake lemma.
But the problem is that $\varphi_1 - \varphi_2$ is not a morphism of extensions.
 A: $\newcommand\mc{\mathcal}\DeclareMathOperator{\Mor}{Mor}\DeclareMathOperator{\Ext}{Ext}$The difference $\varphi_1 - \varphi_2$ need not be a morphism of extensions, so the proof works:
Let $$ 0 \to A \xrightarrow{\iota} E \xrightarrow{p} B \to 0 $$ be an extension with extension class $[\mc E] \in \Ext^1(B,A)$.
Suppose $\DeclareMathOperator{\Aut}{Aut}\varphi_1, \varphi_2 \in \Aut([\mc E])$, then $\varphi_1 - \varphi_2 : E \to E$ is an endomorphism of $E$.
For $b \in B$ we can choose a lift $\ell(b) \in E$
and as $\varphi_1(\ell(b))-\varphi_2(\ell(b))$ maps to zero in $B$,
there is an $a \in A$ such that $\varphi_1(\ell(b)) - \varphi_2(\ell(b)) =\iota(a)$.
This map $f : b \mapsto a$ is well-defined: if $\ell(b) + \iota(a')$ is another lift of $b$ for some $a' \in A$, we have $\varphi_1(\ell(b) + \iota(a')) -\varphi_2(\ell(b) + \iota(a')) = \varphi_1(\ell(b)) - \varphi_2(\ell(b))$ because $\varphi_1 \circ \iota = \varphi_2 \circ \iota = \iota$.
Then by hypothesis $f \in \Mor(B,A) = 0$, which implies $0 = \iota(f(b)) = \varphi_1(\ell(b))- \varphi_2(\ell(b))$ for all $b \in B$ so $\varphi_1(e) - \varphi_2(e) = 0$ for all $e \in E$ by surjectivity of $p$.
Therefore $\varphi_1 = \varphi_2$ so $\Aut([\mc E]) = 0$.
