The setting is as follows: $(R,m)$ is a local ring (assume noetherian, complete, if you need) and $\rho\colon G\to \operatorname{Aut}(M)$ is a group representation on the free, finite-rank $R/m^n$-module $M$, for some $n\geq 1$.
Let $\operatorname{End}(M)$ be the adjoint representation on which $G$ acts by conjugation.
To a class $c\in H^1(G,\operatorname{End}(M))$ one can associate (via an infinitesimal deformation) an exact sequence $0\to M\to E\to M\to 0$ of $(R/m^n)[G]$-modules. The class of this extension in $\operatorname{Ext}^1(M,M)$ is well defined (does not depend on the cocycle representing $c$).
Question: What is a map in the inverse direction?
Note, that if $n=1$, every sequence $0\to M\to E\to M\to 0$ splits as $k:=R/m$-vector spaces. And any such splitting $s$ yields a 1-cocycle $g\mapsto g \cdot s \cdot g^{-1}-s$.
But, e.g. for $n>1$ the sequence of $\mathbb{Z}_2/(4)$-modules $0 \to \mathbb{Z}_2/(2)\to \mathbb{Z}_2/(4)\to \mathbb{Z}_2/(2)\to 0$ does not split.
Or am I missing something?
Given an extension $$ 0 \to M \to N \to M \to 0 $$ apply the functor $\text{Hom}_{R/\mathfrak{m}^n}(M, -)$ to get $$ 0 \to \text{End}(M) \to \text{Hom}(M, N) \to \text{End}(M) \to 0 $$ (using freeness to get exactness on the right). Now take $G$-invariants to get a map $$ \text{End}_G(M) \to H^1(G, \text{End}(M)). $$ The image of the identity endomorphism under this map is your desired cocycle.