Excuse the potato quality drawings below.
First image: In a graph where the distance from one to infinity one equal to the distance from one to the infinitesimal, how would normal shapes that are drawn on the graph below translate in the normal coordinate plane?
The second image for further explanation: If you were to take a square and draw it onto this type of graph where the points of the square go to the infinite points (see bottom image) how would this look translated into the normal coordinate plane, would it match a function we would recognize? What if you did this for a circle or a five-point star?
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:
You can graph this on your favorite graphing site (such as Desmos or Wolfram Alpha), but it will look a little weird.