Fix $n\geq 3$. Does there exist an injective function $f:\mathbb{R}^2\rightarrow\mathbb{R}^2$ such that the image of any circle in $\mathbb{R}^2$ under the map $f$ is a (simple) $n$-gon in $\mathbb{R}^2$?
A friend of mine showed me this problem a few days ago without an answer, and we have been trying to answer it for the better part of 3 days with no luck. We don't really have an intuition about whether or not a function like this exists, but we have managed to prove some partial results.
If we do not require that $f$ is an injection then the answer is positive.
Let $g:\mathbb{R}^2\rightarrow\mathbb{R}$ be the projection map $g(x,y)=x$ and let $P$ be any $n$-gon in $\mathbb{R}^2$. Since $|P|=|\mathbb{R}|=|\mathbb{R}/\mathbb{Q}|$, we can choose a bijection $\varphi:\mathbb{R}/\mathbb{Q}\rightarrow P$, and define $h:\mathbb{R}\rightarrow P$ by $h(x)=\varphi(x+\mathbb{Q})$. If we define $f=h\circ g$, then $f(C)=P$ for every circle $C\subseteq\mathbb{R}^2$.
If we require that $f$ is continuous and injective then the answer is negative.
Assume $f$ exists for a contradiction. Let $C$ be a circle in $\mathbb{R}^2$, and let $p\in\mathbb{R}^2$ such that $f(p)$ lies on an edge (not a vertex) of $f(C)$. Since $f$ is continuous, then its restriction $f|_C$ to the closed interior of $C$ is uniformly continuous. Take a sequence $D_m$ of circles in the tangent to $C$ at $p$ in the interior of $C$ so that $D_m$ converges to $C$ (in the Hausdorff metric). Since $f|_C$ is uniformly continuous, then $f(D_m)$ converges to $f(C)$. Since $f(D_m)$ and $f(C)$ are $n$-gons, it's clear that $\text{vertex}(f(D_m))$ converges to $\text{vertex}(f(C))$. Since each $D_m$ is tangent to $C$ at $p$, then $f(D_m)\cap f(C)=f(D_m\cap C)=\{f(p)\}$, so since $f(p)$ is not a vertex of $f(C)$ then $f(p)$ must be a vertex of $f(D_m)$, since otherwise $f(D_m)\cap f(C)$ would be infinite. The fact that $f(p)\in \text{vertex}(f(D_m))$ for all $m$ contradicts the fact that $\text{vertex}(f(D_m))$ converges to $\text{vertex}(f(C))$.
We also had a proof that there is no bijective $f$, but I may have found an error as I was typing this. Anyways, this brings us no closer to figuring out our actual problem. If anyone has any ideas, we would be happy to hear them.
There is in fact no such injection $f:\mathbb{R}^2\rightarrow\mathbb{R}^2$, which (interestingly enough) we may prove using graph theory. Due to the complexity of the construction here, to hopefully provide a more clean exposition, I have made the choice to phrase things more geometrically, and leave out a few small details in the proof that should hopefully be easy to fill in for anyone interested in doing so.
Choose an integer $m\geq 3$, and assume for contradiction's sake that there exists an injective map $f:\mathbb{R}^2\rightarrow\mathbb{R}^2$ mapping circles to $m$-gons.
It is easy to see that if two circles $C,D$ are tangent at a point $x$, then at least one of $f(C),f(D)$ must have $f(x)$ as a vertex; if $f(C)$ has $f(x)$ as a vertex, we say $C$ stabs $D$ at $x$. It is also easy to see that a circle may stab at most $m$ circles at distinct points.
Let $n\geq 3$. Now we construct a useful two index family $A_{n,k}$ of collections of colored circles:
Let $A_{n,1}$ be the collection consisting of the blue unit circle $U$ at $0$, a ring of $n$ identical red circles (each tangent to their two red neighbours) inscribed in $U$, and a smaller blue circle tangent to each of the red circles. Now, we recurse.
To create $A_{n,k}$ we replace each of the $n$ red circles $C_i$ in $A_{n,1}$ with smaller copies of $A_{n,k-1}$ each rotated appropriately by an angle $\theta_i$ so that this collection has has $n$ rotational symmetries, and for each of the $n^{k+1}$ red circles in this new collection, adding the two unique blue circles with center $0$ tangent to it (ignoring repeats); here we choose the angles $\theta_i$ so that $A_{n,k}$ does not contain three circles that are pairwise tangent at the same point (the fact that we can do this can be proven relatively easily via a cardinality argument). Depicted below are $A_{n,k}$ for a few small values of $n,k$:
We can imbue our collection of circles $A_{n,k}$ with a simple directed graph structure by defining a directed graph $G_{n,k}=(A_{n,k},E_{n,k})$ where $E_{n,k}$ is a set contianing every edge $CD$ such that $C$ stabs $D$ but $D$ does not stab $C$, and one edge out of every pair $CD,DC$ where $C$ and $D$ mutually stab eachother. Note that all information about the graph $G_{n,k}$ except for edge direction is encoded by tangency relations between the circles/vertices in $A_{n,k}$. Now, let $e_{n,k}$ and $v_{n,k}$ be the number of edges and vertices of $G_{n,k}$ respectively. For every vertex $C$ in $A_{n,k}$, we know that $C$ may stab at most $m$ other vertices in $A_{n,k}$ (since no three circles in $A_{n,k}$ are pairwise tangent at the same point), so $\text{out}(C)\leq m$ (this is the outdegree of $C$ in $G_{n,k}$). The degree sum formula then gives us that: $$e_{n,k} = \sum_{C\in A_{n,k}} \text{out}(C) \leq mv_{n,k}$$
Given how $A_{n,k}$ was constructed recursively, it is not too hard to see that $e,v$ satisfy the following recurrence inequalities: \begin{equation}\begin{split} v_{n,k+1} &\leq 2+nv_{n,k}+2n^k\\ e_{n,k+1} &\geq 3n+ne_{n,k}+2n^{k+1}\\ \end{split}\qquad\begin{split} v_{n,1}&=2+n\\ e_{n,1}&=3n\\ \end{split}\end{equation}
Applying induction on $k$, we may deduce the following asymptotic inequalities for $e,v$ with respect to $n$ (fixed $k$): \begin{equation}\begin{split} v_{n,k} &\leq n^k[1+o(1)]\\ e_{n,k} &\geq (2k+1)n^k[1+o(1)]\\ \end{split}\end{equation} Therefore, letting $2k+1>m$, for some large enough value of $n$, we will have that $e_{n,k}>mv_{n,k}$, which is a contradiction. Therefore our injective map $f$ cannot exist.