Let $K$ be an algebraically closed complete non-archimedean field and consider the unit ball $\mathbb{B}^{1}_{K}=Sp(K\langle t\rangle)$. We have an action of $\mathbb{Z}/2\mathbb{Z}$ on $\mathbb{B}^{1}_{K}$ given by sending the non-trivial element in $\mathbb{Z}/2\mathbb{Z}$ to the unique automorphism of $K\langle t\rangle$ which sends $t$ to $-t$. Denote this automorphism by $\xi:\mathbb{B}^{1}_{K}\rightarrow \mathbb{B}^{1}_{K}$. We know that the points in $\mathbb{B}^{1}_{K}$ are given by elements in $K^{\circ}$, and the action of $\xi$ sends a point $\lambda\in\mathbb{B}^{1}_{K}$ to $-\lambda\in\mathbb{B}^{1}_{K}$. Thus, the only fixed point of $\xi$ is the origin. In other words, consider the cartesian diagram: it follows that $Y$ is just a point, which corresponds to the origin. Now let $r$ be the functor which sends rigid analytic $K$-varieties to tfp adic spaces over $Spa(K,K^{\circ})$. Namely, $r(Sp(A))=Spa(A,A^{\circ})$ for any affinoid algebra $A$. According to 1.1.13 page 43 in "Étale cohomology of rigid analytic varieties and adic spaces" the functor $r$ commutes with fiber products. Hence, the fixed locus of the automorphism $\xi$ when regarded as an automorphism of $r(\mathbb{B}^{1}_{K})$ should just be $r(Y)=Spa(K,K^{\circ})$ which is just a classical point in $r(\mathbb{B}^{1}_{K})$. However, in example 2.20 in "Perfectoid spaces", all adic points in $r(\mathbb{B}^{1}_{K})$ are classified. It is clear that the action of $\xi$ fixes all points of type $(2)$ and $(3)$ centered arround the origin. For example, the Gauss point sends a power series $f=\sum a_{n}t^{n}$ to $max\{\vert a_{n}\vert \}$. As $\xi(f)=\sum (-1)^{n}a_{n}t^{n}$, it is clear that the Gausspoint is a fixed point of the morphism $\xi:r(\mathbb{B}^{1}_{K})\rightarrow r(\mathbb{B}^{1}_{K})$. I would like to understand what is wrong about my calculations.

Context: I would like to know if it is true that if an action of a finite group $G$ on an affinoid $K$-variety is free (in the sense that any $g\in G$ which fixes a point in $X$ must be the identity) then the induced action on $r(X)$ is also free (in the same sense as above). My idea for this is that the action is free if and only if for every $g\in G$ the space $X^{g}$ (constructed as in the diagram above ) is empty. As the functor $r$ respects fiber products then if $X^{g}$ is empty it should follow that $r(X^{g})$ is also empty, hence the action of $G$ on $r(X)$ should be free. I don't see any problems with this argument. However, this seems to contradict the example above. Namely, the action of $\mathbb{Z}/2\mathbb{Z}$ on $\mathbb{B}^{1}_{K}$ is not free, but it is when restricted to the boundary $Sp(K\langle t,t^{-1}\rangle)$. However, the induced action on $r(Sp(K\langle t,t^{-1}\rangle))$ fixes the Gauss-point as shown above.

*Scholze, Peter*, **Perfectoid spaces**, Bonn: Univ. Bonn, Mathematisch-Naturwissenschaftliche Fakultät (Diss.). 51 p. (2011). ZBL1296.14020.

*Huber, Roland*, Étale cohomology of rigid analytic varieties and adic spaces, Aspects of Mathematics. E30. Wiesbaden: Vieweg. x, 450 p. (1996). ZBL0868.14010.