A similarity of $\Bbb Q^2$ without a fixed point It is well known that any contraction of $\Bbb R^2$ has a fixed point. In particular, every similarity with the constant different from $1$ has a fixed point. The proof makes use of Banach fixed-point theorem, and hence doesn't translate to $\Bbb Q^2$. Indeed, there are some (non-similarity) contractions of $\Bbb Q^2$ which have no fixed point in it.
Hence my question is:

Are there some similarities of rational plane $\Bbb Q^2$ with constant different from $1$ which have no fixed points?

For completeness, I define a similarity with constant $k$ to be a bijection $S$ of a set $\Bbb Q^2$ such that $d(S(x),S(y))=kd(x,y)$for all points $x,y$, where $d$ is Euclidean distance.
Thanks in advance
 A: A similarity can be described as
$$\begin{pmatrix}x\\y\end{pmatrix}\mapsto
\begin{pmatrix}a&-b\\b&a\end{pmatrix}\cdot
\begin{pmatrix}x\\y\end{pmatrix}+
\begin{pmatrix}c\\d\end{pmatrix}$$
If it maps $\mathbb Q^2$ onto itself, then in particular the origin must be mapped to $\mathbb Q^2$ so $c,d\in\mathbb Q$. With the same argument, since $(1,0)$ maps to $\mathbb Q^2$ then $a+c,b+d\in\mathbb Q$. Together this implies $a,b,c,d\in\mathbb Q$, so this is a rational transformation.
To find its fixed point, simply solve the equation:
\begin{align*}
\begin{pmatrix}a&-b\\b&a\end{pmatrix}\cdot
\begin{pmatrix}x\\y\end{pmatrix}+
\begin{pmatrix}c\\d\end{pmatrix}&=
\begin{pmatrix}x\\y\end{pmatrix}\\
\begin{pmatrix}a-1&-b\\b&a-1\end{pmatrix}\cdot
\begin{pmatrix}x\\y\end{pmatrix}&=
\begin{pmatrix}-c\\-d\end{pmatrix}\\
\end{align*}
This is a linear system of equations with rational coefficients, so if it has a unique solution, that unique solution has to be rational as well. And it will have a unique solution unless $a=1,b=0$, corresponding to translations which are excluded by your $k\neq1$ constraint.
So no, you can't have a similarity which fixes $\mathbb Q^2$ as a whole but doesn't have a fixed point there.
