Let the origin $D'(0,0), C'D' $ the x-axis, $ A'D'$ the y-axis. Let the new origin $D(x_{0},y_{0}), CD $ the new x-axis, $AD $ the new y-axis. A) First, a translation of $A'B'C'D'$ to the point $D $: \[ \left[\begin{array}{clrr}x \\ y \end{array}\right]=\left[\begin{array}{clrr} x_{0} \\ y_{0} \end{array}\right]+\left[\begin{array}{clrr}x' \\y' \end{array}\right]\] Then, an homothety of the translated square into a smaller square, ratio $k$: \[ \left[\begin{array}{clrr}x \\ y \end{array}\right]=\left[\begin{array}{clrr} x_{0} \\ y_{0} \end{array}\right]+\left[\begin{array}{clrr} k & 0 \\ 0 & k \end{array}\right]\left[\begin{array}{clrr}x'' \\y'' \end{array}\right]\] At least, a rotation of this small square into the square $DCBA$, angle $\theta$: \[ \left[\begin{array}{clrr}x \\ y \end{array}\right]=\left[\begin{array}{clrr} x_{0} \\ y_{0} \end{array}\right]+\left[\begin{array}{clrr} k & 0 \\ 0 & k \end{array}\right]\left[\begin{array}{clrr} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{array}\right]\left[\begin{array}{clrr}x''' \\y''' \end{array}\right]\] Calculating: \[\left\{\begin{array}{ll}x=x_{0}+k\cos\theta x'''-k\sin\theta y''' \\ y=y_{0}+k\sin\theta x'''+k\cos\theta y''' \end{array}\right.\] Point $O'(x,y)=O(x''',y''')$: \[\left\{\begin{array}{ll}x=\frac{x_{0}-k(x_{0}\cos\theta+y_{0}\sin\theta)}{k^{2}-2k\cos\theta+1} \\ y=\frac{y_{0-k(-x_{0}\sin\theta+y_{0}\cos\theta)}}{k^{2}-2k\cos\theta+1} \end{array}\quad \quad (1)\right.\] B) Complex numbers. Let $a=x_{0}+i \cdot y_{0} , c = k (\cos\theta+i \cdot \sin\theta)$. Then the function $f(z) = c \cdot z + a $ determines a multiplication with $k$, a rotation with angle $\theta$ and a translation with $a$. $f(z) = z = c \cdot z + a \rightarrow z(1-c) = a \rightarrow z=\frac{a}{1-c} \quad \quad (2)$. (Calculating (2): $x+ i \cdot y = \frac{x_{0}+i \cdot y_{0}}{1-k \cos\theta-i \cdot k\sin\theta}$ $x+ i \cdot y = \frac{(x_{0}+i \cdot y_{0})(1-k \cos\theta+i \cdot k\sin\theta)}{(1-k\cos\theta)^{2}+k^{2}\sin^{2}\theta}$, $x$ and $y$:see (1) ) C) The construction of this point (2)$z=\frac{a}{1-c} $ follows the classical constructions of complex numbers. First, if given $a$ and $c$, construct $1-c$, then $\frac{1}{1-c}$ and at least the product $a \cdot \frac{1}{1-c}$. If wanted, let me know.

Let $T$ be point of intersection $AA'$ and $BB'$ and let $S$ be second intersecton point of circumcircles of trinagles $ABT$ and $A'B'T$. Let $\mathcal{S}$ be spiral similarity with center at $S$ and which takes $A$ to $A'$ and $B$ to $B'$. Then this $\mathcal{S}$ takes $ABCD$ to $A'B'C'D'$ and $S$ is ''obviously'' unique fixed point. So $O=S$.

easy Draw a line parallel to $AB$ on the small map. Draw the corresponding line on the big map. As the lines vary, their intersection will line on a line $\l_1$. Do the same with $BC$. There will be another line $l_2$. $O$ is the intersection of $l_1$ and $l_2$ Here is a construction with just a straight-edge:

You can use this lemma. $AB\cap A'B'=M$ $CD\cap C'D'=N$ Call $X$ the intersection of an arbitrary line parallel to $AB$ and the corresponding line (parallel to $A'B'$). $M,X,N$ are collinear.