In terms of intuition on why these polygons exist, we note that a circle is just a one-petaled rose. We expect the intersection points of a one-petaled rose and the given $n$ petaled-rose of the same petal length to have some periodic structure (about $\theta$). So if some subset of the intersection points periodically go around the one-petaled rose, they are therefore evenly distributed around a circle, and will form a regular polygon.
To rigorize this, we will now completely classify all regular $k$-gons that can be contained in a rose curve $r=\cos n\theta$ under the assumption that its circumcircle passes through the origin and has diameter $1$.
Lemma: Consider a circle $C$ passing through the origin, and a set of points $P=\{P_1,P_2,\dots,P_k\}$ on $C$, each with argument $\theta_i=\arg P_i$. Then, $P$ forms a regular $k$-gon if and only if $\{\theta_1,\dots,\theta_k\}$ forms an arithmetic progression with difference $\frac{\pi}{k}$ (modulo $2\pi$).
Proof: For the forward implication, WLOG let the points lie on the circle counterclockwise in the order $P_1,\dots,P_k$, and let the origin $O$ lie on the minor arc between $P_1$ and $P_k$. For $1\leq i\leq k-1$, the angle $\angle P_iOP_{i+1}$ sees arc $P_iP_{i+1}$, which has measure $\frac{2\pi}{k}$ since $P$ forms a regular polygon. Thus, $\theta_{i+1}-\theta_{i}\pmod {2\pi}=\angle P_iOP_{i+1}=\frac{\pi}{k}$. This means $\theta_1,\dots,\theta_k$ form an arithmetic progression with difference $\frac{\pi}{k}$, as desired.
For the reverse implication, WLOG let the arithmetic progression be $\theta_1,\dots,\theta_k$ in that order. Using the same logic as before, $\angle P_iOP_{i+1}=\frac{\pi}{k}$, so arc $P_iP_{i+1}$ has measure $\frac{2\pi}{k}$. This means the points are all equally spaced around $C$, so $P$ forms a regular polygon. $\square$
Call a number $k\geq 3$ valid for a given $n$ if a regular $k$-gon can be contained in the rose curve $r=\cos n\theta$ with its circumcircle crossing the origin and having diameter $1$. We claim the following:
- When $n$ is odd, $k$ is valid if it divides $\frac{n-1}{2}$ or $\frac{n+1}{2}$.
- When $n$ is even, $k$ is valid if divides of $n-1$ or $n+1$.
It suffices to consider the circle $r=\cos\theta$ and its intersections with the rose curve. The intersection points occur when $\cos(\theta)=\cos(n\theta)$. This means $\pm\theta+2\pi b=n\theta$ for $b\in\mathbb{Z}$, so $\theta=\frac{2\pi b}{n\mp 1}$.
If $n$ is odd and $\theta=\frac{2\pi b}{n-1}$, then $b$ and $b+\frac{n-1}{2}$ are the same point since $\theta$ gets increased by $\pi$ and $r=\cos\theta$ gets negated. Thus, the distinct solutions are captured when $b$ ranges from $0$ to $\frac{n-3}{2}$. When $b>\frac{n-1}{4}$, the radius becomes negative, so the argument of those points becomes $-\pi+\frac{2\pi b}{n-1}$. So we get an arithmetic sequence for the arguments starting with the solution when $b=\left\lceil\frac{n-1}{4}\right\rceil$ and going up by a difference of $\frac{2\pi}{n-1}$. From our lemma, this creates a regular $\frac{n-1}{2}$-gon. Similarly, for the $\theta=\frac{2\pi b}{n+1}$ case, we get a regular $\frac{n+1}{2}$-gon.
If $n$ is even and $\theta=\frac{2\pi b}{n-1}$, then all solutions when $b$ ranges from $0$ to $n-2$ are distinct, since none differ by $\pi$ or $2\pi$. When $\frac{n-1}{4}<b<\frac{3(n-1)}{4}$, the radius becomes negative, so we actually get arguments of $\frac{2\pi b}{n-1}+\pi=\frac{(2b-1)\pi}{n-1}$ (modulo $2\pi$, noting that $n$ is even). Putting all this together, the arguments of all these intersection points starts from $-\frac{n-2}{2(n-1)}\pi$ and increases by $\frac{\pi}{n-1}$ (the negative radii points effectively fill in the halfway gaps for the positive points, leading to an increase of $\frac{\pi}{n-1}$ as opposed to the odd case). From our lemma this forms a regular $(n-1)$-gon. Similarly, when $\theta=\frac{2\pi b}{n+1}$, we get a regular $(n+1)$-gon.
If $k$ is valid, then the divisors of $k$ will be trivially valid by selecting the points of $k$-gon periodically. Our claim is now proved.
Claim: Any origin-intersecting circle $r=\cos(\theta-a)$ with $a\in[0,2\pi)$ creates regular $k$-gon through its intersection points with $r=\cos n\theta$ if and only if $r=\cos\theta$ also does.
We can rewrite the solution cases to $\cos(\theta-a)=\cos n\theta$ as the positive case $\theta+\frac{a}{n-1}+2\pi b=n\theta+\frac{na}{n-1}$ and negative case $-\theta-\frac{a}{n+1}+2\pi b=n\theta+\frac{na}{n+1}$. This means for each solution $\theta_p$ to the positive case of $\cos\theta=\cos n\theta$, we know $\theta_p+\frac{a}{n-1}$ is a solution to the positive case of $\cos(\theta-a)=\cos n\theta$. Similarly, a solution $\theta_n$ to the negative case of the $r=\cos\theta$ circle will correspond to a solution $\theta_n+\frac{a}{n+1}$ to the negative case here.
Thus, all the arguments will be shifted by the same constant amount for each polygon in the $r=\cos\theta$ case, meaning they still form an arithmetic progression with preserved difference. Our lemma still applies, and we get the desired regular $k$-gons. For the other direction, we can just shift the solutions back with the same logic.
Because our characterization of the intersection points for $\cos\theta=\cos n\theta$ was complete, our characterization for $\cos(\theta-a)$ for all $a\in [0,2\pi)$ is also complete.
To summarize:
There is an infinite family of regular $k$-gons contained in the rose curve when $k$ divides $\frac{n-1}{2}$ or $\frac{n+1}{2}$ for odd $n$, and when $k$ divides $n-1$ and $n+1$ for even $n$. Moreover, these are the only regular $k$-gons contained in the rose curve with a circumcircle of diameter $1$ that passes through the origin.
I conjecture that the regular $k$-gons above, in addition to the trivial case when $k\mid p(n)$, are the only ones contained in any rose curve $r=\cos n\theta$. Proving this seems very tedious, but I suppose a first step would be to show this for polygons with circumcircles passing through the origin but with radius less than $1$.
