Suppose $\Gamma$ is a finite group acting on another group $A$. If $A$ is abelian, then it follows from basic group theory that the cohomology classes $[c] \in H^1(\Gamma, A)$ are in one-to-one correspondence with each other, for instance under the map $$[c] \rightarrow [d], c' \mapsto c'-c+d.$$ Roughly speaking, all cohomology classes have the "same size."
If $A$ is nonabelian, we can still construct the pointed set $H^1(\Gamma, A)$ by constructing the set of 1-cocycles $$Z^1(\Gamma, A) = \{ c : \Gamma \rightarrow A : c(\gamma\gamma') = c(\gamma) \gamma.c(\gamma') \text{ for all } \gamma, \gamma' \in \Gamma \},$$ and declaring $H^1(\Gamma, A)$ to be the set of equivalence classes in $Z^1(\Gamma, A)$ under the relation $$c \sim d \Leftrightarrow \text{there exists } a \in A \text { such that } c(\gamma) = a^{-1} d(\gamma)(\gamma.a) \text{ for all } \gamma \in \Gamma.$$ If $A$ is abelian, this definition coincides with the usual case.
My question is: is it true in the general case that there is such a correspondence between the elements of the classes in $H^1(\Gamma, A)$?
I believe that all of the cohomology classes in $H^1(\Gamma, A)$ should be the "same size", but I'm having trouble proving or disproving this. If it's relevant, in my case $\Gamma$ is a Galois group acting on a nonabelian algebraic group.
I think the answer to this question is no in generality, consider the integers $\mathbb{Z}$ acting trivially on a finite group $G$. The $1$ cocycles are sections of the semi direct product projection, which are just elements of $G$ since the action is trivial. The relation which yields cohomology classes is conjugation, and so $H^1(\mathbb{Z},G)$ is just the conjugacy classes of $G$, and these won’t all be the same size if $G$ isn’t abelian.