Is $x\mapsto (ax+b)/2^n$ in the affine groups of $\Bbb Z_2^\times$, $\Bbb Z_2$ and $\Bbb Q_2$?
Or what quantification of $a,b, n$ is necessary for $f$ to be the group of invertible affine transformations?
My Attempt
So I think for $\textrm{Aff}(\Bbb Z_2^\times)$ I need the bijections $f:\Bbb Z_2^\times\to\Bbb Z_2^\times$ in which case I need $a\in\Bbb Z_2^\times, b\in\Bbb Z_2\setminus\Bbb Z_2^\times, n=0$
In $\textrm{Aff}(\Bbb Z_2)$ it's less clear to me but there's more flexibility . I think I need the injections $f$ such that both $f$ and $f^{-1}$ are onto $\Bbb Z_2$. Tentatively therefore I think that I need $(ax+b):a\in\Bbb Z_2^\times, b\in \Bbb Z_2^\times$, correct?
Then in $\textrm{Aff}(\Bbb Q_2)$ it looks like there's a lot more freedom, in fact anything of the form $(ax+b)/2^n:a\in\Bbb Z_2^\times, b\in\Bbb Q_2, n\in\Bbb Z$.
Is that correct? I feel like there should be a general rule for the affine transformations over the units, ring and field which instantly answers this question.