Does this inversion exist? Is there an inversion that takes the points (2,0), (−2,0), (0,2), (0,−1) into
vertices of a square?
The question seems very straightforward and we can see if there is a circle at the origin, (0,0) with radiuse $\sqrt{2}$ then point (0,-1) will invert to (0,-2) and the points can be a square. or vice versa regarding the other three points. My question is: is it possible to consider points in isolation like this regarding the question? Or would it be more sensible to say since inversion maintain distances there is no possible inversion on all points to make this a square?
 A: The answer is no. In fact, any inversion maps a circumference to a circumference or a line. Therefore, its inverse also. That is to say, if there was an $f$ that fulfills the request, then its inverse should send the circumcircle of the square in the circle with center $(0,0)$ and radius 2. But the point $(0,-1)$ is not there.
A: In the plane, circle inversion is a conformal (angle-preserving) mapping, not a distance-preserving mapping.  Therefore, it is immediately obvious that a mapping that satisfies your criteria does not exist, because the angle at the vertex $(0,-1)$ is not $90^\circ$, and no inversion can change this.
A: The inversion with respect to the circle with center $z_0$ and radius $R$ is given by $I(z)=\frac{R^2}{\overline{z}-\overline{z_0}}+z_0$ and it is not a Mobius transformation.
Let $A',B',C',D'$ be the images of $A=2, B=2i, C=-2, D=-i$ respectively under an inversion $I(z)$ with center $z_0$. We want the quadrangle $\square A'B'C'D'$ to be a square. Due to the symmetry with respect to $y$-axis, or from the equation $A'B'=B'C'$, $z_0=iy$ is pure imaginary. Furthermore, by solving the equation $A'B'=C'D'$ we find that $z_0=(2\sqrt{10}+6)i$ or $(2\sqrt{10}-6)i$.
Unfortnately, in both cases, although we obtained a quadrilateral (equilateral quadrangle), the inner angles are not right.
