I have been reading MacLane's book on Homology and he defines a group extension to be a short exact sequence
$$ 0 \xrightarrow{x} G \rightarrow B \xrightarrow{\sigma} C \rightarrow 1 $$
So far so good. But then he says this wich i quite dont understand what is going on.
Conjugation in $B$ yields a homomorphism $\theta : B\to \operatorname{Aut}(G)$ such that $x[(\theta b)g]=b+(xg)-b$. Suppose $G=A$ abelian then $\theta(A)=1$ so it induces an homomorphism $\phi : C \to \operatorname{Aut}(A)$ with $\phi \sigma = \theta $.
I really dont understand this last paragraph, I dont get what he means with conjugation in $B$ and is he identifying $A$ with $x(A)$ ? Also I don't get how he pulls that last homomorphism maybe from the fact that its a short exact sequence?
Any help is appreciated, thanks in advance.
Correcting your formula, and using $A$ from the start, consider the short exact sequence $$1 \to A \xrightarrow{x} B \xrightarrow{\sigma} C \to 1 $$
"Conjugation in $B$" is a shorthand for the homomorphism $I : B \to \text{Aut}(B)$ defined by requiring $I(b)$ to be the the "conjugating" map $$I(b)(c) = b c b^{-1} $$ which is an automorphism of $B$.
The homomorphism $x$ is injective, and it is very common to identify the domain of an injective function with its image, so yes, he is identifying $A$ with its image under $x$ which is a normal subgroup of $B$.
The fact that $A$ (or, if you like, $x(A)$) is a normal subgroup of $B$ implies that $I(b)(A)=A$ for all $b \in B$. Therefore, each $I(b) : B \to B$ restricts to an automorphism of $A$ denoted $\phi(b) : A \to A$. This defines a function $\phi : B \to \text{Aut}(A)$. Since $I : B \to \text{Aut}(B)$ is a homomorphism, it follows that $\phi$ is also an homomorphism.