Extend Group Action of $\mathbb{A}^1 /G$ to Projective Line My question refers to an argument used in Torsten Ekedahl's paper: https://arxiv.org/abs/0903.3148
in Example ii) (page 8):
We consider a finite subgroup of affine transformation of $\mathbb{A}^1$. The main goal is to show that the class of the classifying stack $\{BG\}$ is $1$ in $K_o(Stck_k)$ (the terminology is explained at the beginning). 
Here the excerpt:

My question is why and how to see that for the qotient scheme we have $\mathbb{A}^1 /G \cong \mathbb{A}^1$.
Using the canonical open embeeding $\mathbb{A}^1 \subset_o \mathbb{P}^1$ as affine chart I tried to reduce the problem to the case $\mathbb{P}^1/G$ via following diagram (I'm not sure if $f$ is well defined):
$$
\require{AMScd}
\begin{CD}
\mathbb{A}^1 @>{i}  >> \mathbb{P}^1  \\
@VVprV  @VVprV  \\
\mathbb{A}^1/G @>{f}>> \mathbb{P}^1/G 
\end{CD}
$$
Here occure two questions:
Can I here extend the $G$-action to the projective line canonically?
And if yes, is it esier to show that $\mathbb{P}^1 /G \cong \mathbb{P}^1$ in this case & how?
 A: I am writing this as an answer.
Lemma. For every perfect field $k$, an integral, affine $k$-curve $U_k$ is $k$-isomorphic to $\mathbb{A}^1_k$ if and only if it is normal and unirational whose natural homomorphism of unit groups, $\overline{k}^\times \to \overline{k}[U_{\overline{k}}]^\times,$ is an isomorphism.
Proof.  The "only if" statement is immediate; we prove the "if" statement.  For every field $k$, every separated, connected, normal, finite type $k$-scheme of dimension $1$ admits a dense open immersion into a projective, connected, normal $k$-scheme of dimension $1$, and this is unique up to unique isomorphism.  If $k$ is perfect, then the projective $k$-scheme is even $k$-smooth.  
A connected, smooth, projective $k$-curve is (geometrically) rational if it is (geometrically) unirational.  Thus, every unirational, normal $U_k$ is isomorphic to a dense open in $\mathbb{P}^1_k$.  If the complementary reduced divisor $D_k$ has degree $r+1$, then the base change $D_{\overline{k}}\subset \mathbb{P}^1_{\overline{k}}$ consists of $r+1$ closed points.  By  straightforward computation, the quotient group $\overline{k}[U_{\overline{k}}]^\times/\overline{k}^\times$ is a free Abelian group of rank $r$.  Thus, if this quotient group is trivial, then $D_k$ has length $1$, i.e., $D_k$ is a $k$-point.  Up to a $k$-automorphism of $\mathbb{P}^1_k$, assume that this $k$-point is the point at infinity.  Thus, $U_k$ is isomorphic to $\mathbb{A}^1_k$. QED 
Now the comments above finish the proof.  The geometric quotient $U_k$ of $\mathbb{A}^1_k$ by a finite group $G$ of automorphisms is an affine, connected, normal $k$-group.  Since every unit of $U_k$ pulls back to a unit of $\mathbb{A}^1_k$, it follows that the unit group of $U_k$ equals the constants.  Thus, $U_k$ is $k$-isomorphic to $\mathbb{A}^1_k$.
