Automorphisms of a certain variety Which is the group of automorphisms of the variety $(\mathbb{A}^1\setminus\{0,1\})^{\times 3}$ ? 
($0,1$ are two points of $\mathbb{A}^1$, say the points corresponding to $0,1$ of the base field $\Bbbk$, under the identification of the closed points of $\mathbb{A}^1$ with $\Bbbk$)
 A: First suppose for simplicity that $k$ has characteristic $0$. We have the following 

Lemma: Let $K$ be a field and let $f$ be a non-constant unit of $K[t, \frac{1}{t}, \frac{1}{1-t}]$ such that $f-1$ is also a unit, then 
  $$f\in \left\{ t, 1-t, \frac{1}{t}, \frac{1}{1-t}, \frac{t}{t-1}, \frac{t-1}{t}\right\}. $$

Let $S=\mathbb A^1_k \setminus \{ 0, 1 \}$. Let's see how to determine the automorphism group of $S^n$ for any positive integer $n$ using this lemma. 
Note that 
$$O(S)=k[t_1, t_2,...,t_n, \frac{1}{t_1}, ..., \frac{1}{t_n}, \frac{1}{1-t_1},..., \frac{1}{1-t_n}]$$ 
and $O(S)^*=k^*t_1^{\mathbb Z}...t_n^{\mathbb Z}(1-t_1)^{\mathbb Z}...(1-t_n)^{\mathbb Z}$. An automorphism of $S^n$ induces an automorphism $f: O(S)\to O(S)$. In particular $f(t_1)\in O(S)^*$ and $f(t_1)-1=f(t_1-1)\in O(S)^*$. As $f(t_1)$ is non-constant, there exists $j\le n$ such that $f(t_1)$ is non-constant in $t_j$ (if $k$ has positive characterisitc $p$, we can require that $f(t_1)$ is not a $p$th power in $t_j$). Let $K=k(t_r)_{r\ne j}$. Then $f(t_1)$ and $f(t_1)-1$ are both non-constant units in $K[t_j, 1/t_j, 1/(1-t_j)]$. By the above lemma, $f(t_1)\in \mathbb F(t_j)$ where $\mathbb F$ is the prime field of $K$ (which is also the prime field of $k$). We can repeat the same argument to any $f(t_i)$. We see that there exists a map 
$$\sigma: \{1,...,n\}\to \{1, ..., n\}$$
such that $f(t_i)\in\mathbb F(t_{\sigma(i)})$ for all $i\le n$. As $f$ is an automorphism, $\sigma$ must be surjective, hence bijective. The permutation $\sigma$ acts naturally on $S^n$ by permuting the coordinates. And $f\circ\sigma^{-1}$ maps $t_i$ to an element of $\mathbb F(t_i)$ for all $i\le n$. So $f\circ \sigma^{-1} \in \mathrm{Aut}(S)\times ... \times \mathrm{Aut}(S)=\mathrm{Aut}(S)^n$. Therefore

We have an exact sequence of groups
  $$ 1\to \mathrm{Aut}(S)^n\to \mathrm{Aut}(S^n)\to \mathcal{S}_n \to 1$$
  with a natural section $\mathcal{S}_n \to \mathrm{Aut}(S^n)$, where $\mathcal{S}_n$ is the symetric group on $n$ elements. 

Note that the group $\mathrm{Aut}(S)$ is isomorphic to the symetric group $\mathcal{S}_3$ by the above lemma. 
It remains to prove the lemma. There is a quick geometric way to show $f$ is a homography leaving globally stable the set $\{0, 1, \infty\}$. Then the lemma is a classical result. Let us give a more direct proof. Write
$f(t)=ct^r(1-t)^s$ with $c\in K^*$ and $r, s\in \mathbb Z$. Then there exists $d\in K^*$, $q, \ell\in \mathbb Z$ such that 
$$ct^r(1-t)^s-1=dt^q(1-t)^{\ell}.$$
Suppose $r>0$. Evaluating at $t=0$, we see that $q=0$. So
$$ct^r(1-t)^s-1=d(1-t)^{\ell}.$$
If $s>0$, then $\ell=s+r>0$ and $t-1 | 1$, impossible. If $s=0$, we see that $f(t)=t$ (here we use the fact that $r=\ell$ is non zero in $k$). If $s<0$, we get $s=\ell$ and $f(t)=t/(t-1)$. The remaining cases can be dealed with using the symetries $t\mapsto 1-t$, $t\mapsto 1/t$. 
Final remark: when $k$ has characteristic $p>0$, the lemma still holds if $f$ is not a $p$th power in $K(t)$. The proof is same as above.
