Let me give you a fuller development: There are two planes: The real plane, with coordinates $(x,y)$. That is, $x, y \in (-\infty, \infty)$, the set of all real numbers. Secondly, there is the transformed plane, with coordinates $(x', y')$ which can only take values in $(-2, 2)$. This is the plane in your graphs, though you've labeled the axes with their $x$ and $y$ equivalents. Your square consists of the line segments $$\begin{cases}
y' = 2 - x' & x' \in [0,2), y' \in [0,2)\\
y' = 2 + x' & x' \in (-2, 0], y' \in [0,2)\\
y' = -2 - x' & x' \in (-2, 0], y' \in (-2, 0]\\
y' = -2 + x' & x' \in [0,2), y' \in (-2, 0]\end{cases}$$
(I've assumed you are familiar with interval notation. If not, $(-2, 2) = \{t\mid -2 < t < 2\}, [0,2) = \{t \mid 0 \le t < 2\}, (-2,0] = \{ t \mid -2 < t \le 0\}$. And yes, this is ambiguous with using $(x,y)$ to represent the ordered pair of coordinates. For this reason, some people prefer, for example, $]-2,2[$ instead of $(-2,2)$. But that always looks clunky to me, and I've never known even slow students to have much trouble distinguishing between the two by context.)
To figure out how $(x,y)$ are related to $(x', y')$, look first at the case $x \ge 1$. Your statement "The distance from 1 to n is the same as the distance from 1 to 1/n", I am interpreting to mean that for any $x \ge 1$, you want the distance from $1$ to the corresponding value of $x'$ to be the same as the distance from $1$ to $\frac 1x$. Since I don't think you want the ordering to change, if $x > 1$, then so should be $x'$. Thus the distance from $1$ to $x'$ is $x' - 1$, while the distance from $1$ to $\frac 1x$ is $1 - \frac 1x$. Therefore
$$x' - 1 = 1 - \frac 1x$$
So, $$x' = 2 - \frac 1x\\x = \frac 1{2 - x'}$$
It is not clear to me what you want to occur for $x$ between $0$ and $1$. You could just let $x' = x$ in this region, but the transition at $x = 1$ is not smooth, so nice smoothly varying curves in one plane will have corners at $x = x' = 1$ in the other plane. Presumably for $x < 0$ one should reflect the behavior for $x > 0$. This gives the full transformation:
$$x' = \begin{cases} -2 - \frac 1x & x \le 1\\x & -1 \le x \le 1\\2- \frac 1x & 1 \le x\end{cases}$$
which has inverse
$$x = \begin{cases}\frac 1{-2 - x'} & x' \le 1\\x & -1 \le x' \le 1\\\frac 1{2 - x'} & 1 \le x'\end{cases}$$
The exact same relationship holds between $y$ and $y'$. So if I look at the case where $x' > 1$ and $y' > 0$, the equation is $y' = 2 - x'$, which gives $y' < 1$, so $y = y'$, while $x' = 2 - \frac 1x$, and the equation becomes
$$y = 2 - (2 - \frac 1x) = \frac 1x$$
And when $0 \le x' \le 1$, then $y' > 1$ so now $y' = 2 - \frac 1y$ and $x' = x$, and the equation becomes
$2-\frac 1y = 2 - x$, so once again $y = \frac 1x$, as you claimed. (In my now-deleted comment, I extended the $x, x'$ relationship above $1$ to the region between $0$ and $1$ as well, but that doesn't work. I only realized later that you apparently are making no change in that region.)
To consider the circle $x'^2 + y'^2 = 4$, we make the substitutions as before. Again, I'll only look at the first quadrant:
- When $x' \in [1, \sqrt 3], y' \in [1,\sqrt 3]$, we have $$\left(2-\frac 1x\right)^2 + \left(2 - \frac 1y\right)^2 = 4$$
- When $x' \in [\sqrt 3, 2), y' \in (0,1]$, we have $$\left(2 - \frac 1x\right)^2 + y^2 = 4$$
- When $x' \in (0,1], y' \in [\sqrt 3, 2)$, we have $$x^2 + \left(2 - \frac 1y\right)^2 = 4$$
You can graph this on your favorite graphing site (such as Desmos or Wolfram Alpha), but it will look a little weird.
