Perfect circles in the Mandelbrot set? It is known that the boundary of the period 2 hyperbolic component of the Mandelbrot set is a perfect circle of radius $\frac{1}{4}$ centered at $-1$.  Moreover it is known that the boundaries of the circle-like period 3 hyperbolic components are not perfect circles ("A Parameterization of the Period 3 Hyperbolic Components of the Mandelbrot Set",  Dante Giarrusso, Yuval Fisher).
Question: is the period 2 component the only hyperbolic component in the Mandelbrot set whose boundary is a perfect circle?
Similarly, is the period 1 component the only hyperbolic component in the Mandelbrot set whose boundary is a perfect cardioid?
I suspect the answer to both questions is "yes", but haven't found any conclusive references.
 A: Here's something I tried, thanks to Adam's comment for the basic idea for $p \ge 3$.  This answer is missing some details, comments suggesting improvements are welcome, as would be other answers that fill in the gaps.  It also relies on the fact (?) that:
$$\forall 0 \neq a \in \mathbb{C}, 0 \neq b \in \mathbb{C}, 1 < m \in \mathbb{N} . \exists x . |a e^{i x} + b e^{i m x}| \neq 1$$

Let $F(z, c) = z^2 + c$ with $F^{p+1}(z, c) = F^p(F(z, c), c)$.  Now the boundary of a hyperbolic component can be parameterized by $\theta \in \mathbb{R}$ by the solution of the equation system:
$$
F^p(z,c) = z \\
\frac{\partial}{\partial z}F^p(z,c) = e^{i \theta}
$$
Now the question reduces to showing $c$ is of the form $c = c_0 + r_0 e^{i \phi}$ where $c_0 \in \mathbb{C}$ and $r_0 \in \mathbb{R}$ are constants and $\phi \in \mathbb{R}$.
$F^p(z,c) = z$ defines a polynomial of even degree $P(z) = 0$, whose constant coefficient is the product of its roots and is a polynomial in $c$ of degree $2^{p-1}$.  The roots include those of $F^q(z, c) = z$ where $q | p$.  Also, $\frac{\partial}{\partial z}F^p(z, c) = 2^p \Pi z_k$ where the $z_k$ are the $p$ roots in the periodic orbit of the desired solution $z$ (all $z_k$ are roots of $F^p(z, c) = z$, the remaining roots have lower period).
Case $p = 1$:
$$ z^2 + c_0 + r_0 e^{i \phi} = z \\
\therefore z = \frac{1 \pm \sqrt{1 - 4(c_0 + r_0 e^{i \phi})}}{2} \\
\frac{\partial}{\partial z} = 2 z = e^{i \theta}
$$
Now $|e^{i \theta}| = 1$ but $\exists x . |2 z| = |1 \pm \sqrt{x}| \neq 1$, so conclude that period $1$ component is not a perfect circle.
Case $p = 2$:
The equations reduce to
$$ 4(1 + c_0 + r_0 e^{i \phi}) = e^{i \theta} $$
with obvious solution $c_0 = -1, r_0 = \frac{1}{4}, \phi = \theta$, so conclude that the period 2 component is a perfect circle.
Case $p = 3$:
The equations reduce to
$$ 8 (c^3 + 2c^2 + c + 1) = e^{i \theta} \text{ where } c = c_0 + r_0 e^{i \phi}$$
For this to hold, the coefficients of $e^{i k \phi}$ must be zero for all $k > 1$.  But setting $k = 3$ implies $r_0^3 = 0$ but we know that $r_0 > 0$ as hyperbolic components have non-empty interior.  Contradiction, conclude that no period 3 component is a perfect circle.
Case $p > 3$:
Similarly to the $p = 3$ case, get a polynomial of degree $m > 1$ in $e^{i \phi}$ whose highest term has coefficient $r_0^m$.  It remains to show that the polynomial really does have degree greater than $1$.  The constant coefficient (product of roots) is a polynomial of degree $2^{p-1}$ in $c$, divided by the corresponding constant coefficient of all smaller divisors of the period gives:
$$m = 2^{p-1} - \sum_{q | p, q < p} 2^{q-1}$$
which solved numerically gives:
$$\begin{aligned}
p & & & m \\
1 & & & 1 \\
2 & & & 1 \\
3 & & & 3 \\
4 & & & 5 \\
5 & & & 15 \\
6 & & & 25 \\
\vdots
\end{aligned}$$
Finally, $m > 1$ for all $p \ge 3$ because $\exists q > 1 . q \nmid p$.
