The short answer is that there is not much to say about the relationship between $H^1(G, B)$ and a twist $H^1(G, B_c)$ where $c$ is a cocycle taking values in $Aut(B)$. (I am going to write $B_c$ for the twist instead of Serre's notation $_cB$ for the sake of easy typesetting.) You can get a good feel for what is possible by soaking in sections I.5.7 and III.1.4 of Serre's *Galois Cohomology*.

## Section I.5.7

One thing you *can* do -- as exhibited in I.5.7 -- is twist all three terms in a short exact sequence of $G$-modules and get a new short exact sequence, assuming the obvious compatibility conditions hold. Serre starts with an exact sequence

$1 \to A \to B \to C \to 1$

where $A$ is assumed central in $B$. Then he fixes a 1-cocycle $c$ with values in $C$ and twists to get an exact sequence

$1 \to A \to B_c \to C_c \to 1$.

Note that this twist $B_c$ is *not* an inner twist of $B$, because $c$ need not be in the image of $H^1(G, B) \to H^1(G, C)$.

This may look like a lame example, in that the twist of $B$ is "pretty close" to being inner. But already here you don't have any results regarding a connection between $H^1(G, B)$ and $H^1(G, B_c)$. That's a pretty fuzzy statement; Serre says as much as you can say with precision in Remark 1: "it is, in general, false that $H^1(G, B_c)$ is in bijective correspondence with $H^1(G, B)$."

## Section III.1.4

This section discusses your question for the specific case where $G$ is the absolute Galois group of a field $k$ and $B$ is the group of $n$-by-$n$ matrices of determinant 1 with entries in a separable closure of $k$. Serre explains what you get as $B_c$ when you twist $B$ by a cocycle with values in $Aut(B)$. You can get, for example, a special unitary group.

You can find explicit descriptions of $H^1(G, B_c)$ for some $B_c$'s in *The Book of Involutions*, pages 393 (Cor. 29.4) and 404 (box in middle of page). Note that for $B_c$ as in Section I.5.7, $H^1(G, B_c)$ is a group (a nice coincidence) but in the case where you get a true special unitary group, $H^1(G, B_c)$ does not have a reasonable group structure--it is just a pointed set like you expect.