Symmetries of Mandelbrot sets with integer exponents I have been experimenting with recursive formulas of the form:
\begin{equation}
\forall c \in \mathbb{C} , z_{n+1} = z_{n}^\alpha + c
\tag{1}
\end{equation}
as well as:
\begin{equation}
\forall c \in \mathbb{C}, z_{n+1} = \overline{z_{n}}^\alpha + c
\tag{2}
\end{equation}
where $z_0 = c$, $\alpha \in \mathbb{Z}$ and $|\alpha| > 1$. 
I made the following observations:


*

*In case $(1)$, the resulting structure has $\alpha-1$ symmetries when $\alpha \geq 2$ and $|\alpha|+1$ symmetries when $\alpha \leq -2$.

*In case $(2)$, the resulting structure has $\alpha+1$ symmetries when $\alpha \geq 2$ and $|\alpha|-1$ symmetries when $\alpha \leq -2$.


If you'd like to experiment with the software I used to gain more insights, I made it publicly available: https://github.com/AidanRocke/TensorFlow-Fractals
Here are a few visualisations of case $(1)$ for $\alpha \in {2,4,-2,-4}$:




So far I don't have an explanation for all four observations but I have a stability argument for case $(1)$ where $\alpha > 1$:
\begin{equation}
z_{n+1} \sim z_0^{\alpha(n+1)} 
\end{equation}
which may be deduced by multiplying and dividing all terms in the series expansion by $z_0^{\alpha(n+1)}$. As a result, near the boundary of the circumscribing disk with radius $R$ the most stable(and therefore most distant) points are near the roots of unity:
\begin{equation}
\mathcal{U}_n = \{e^{\frac{2ik\pi}{\alpha}}: k \in [0,\alpha-1]\}
\end{equation}
I think this argument is sufficient but feel free to correct me if I'm wrong. As for the three other cases, I don't have a good explanation yet. 
 A: Given the observed geometry of these figures, we are looking for reflection and rotation symmetries where the axes of reflection and rotation coincide. Now, a rotation in the complex plane $\mathbb{C}$ is equivalent to multiplication by some $R \in \mathbb{C}$ where $|R|=1$ whereas a reflection in $\mathbb{C}$ about the axis $R=e^{i\theta}$ is equivalent to the transformation $z \mapsto \overline{z}R^2$.


*

*Given the first recurrence relation:


\begin{equation}
\forall c \in \mathbb{C},z_{n+1} = z_n^\alpha + c
\tag{1}\end{equation}
we are looking for $R$ that simultaneously satisfy:
\begin{equation}
\begin{cases}
(Rz_n)^\alpha + Rc=R(R^{\alpha-1}z_n^\alpha + c)=R(z_n^\alpha + c) \\
\lvert(\overline{z_n}R^2)^\alpha+\overline{c}R^2\rvert^2=\lvert \overline{z_n}^{\alpha}R^{2(\alpha-1)}+\overline{c}\rvert^2=\lvert \overline{z_n}^{\alpha}+\overline{c}\rvert^2=
\lvert z_n^{\alpha}+c \rvert^2
\end{cases}
\end{equation}
So we must solve for $R$ where $R^{\alpha-1}=1$ and we find that:
\begin{equation}
R \in \mathcal{U}_{\alpha-1}=\{e^{\frac{2 k\pi i}{\alpha-1}}:k \in [0,..,\alpha-2\}
\end{equation} 
This corresponds to $\alpha−1$ symmetries when $\alpha \geq 2$ and $|\alpha|+1$ symmetries when $\alpha \leq -2$.


*Given the second recurrence relation:


\begin{equation}
\forall c \in \mathbb{C},z_{n+1} = \overline{z_n}^\alpha + c=\left(\frac{z_n^{-1}}{|z_n|^2}\right)^\alpha + c
\tag{2}\end{equation}
we are looking for $R$ that simultaneously satisfy:
\begin{equation}
\begin{cases}
\left(\frac{R^{-1}z_n^{-1}}{|Rz_n|^2}\right)^\alpha + Rc=R\left(\frac{R^{-\alpha-1}z_n^{-\alpha}}{|z_n|^{2\alpha}}+c \right)=R\left(\left(\frac{z_n^{-1}}{|z_n|^2}\right)^\alpha + c\right) \\
\big\lvert \left(\frac{R^{-2}\overline{z_n}^{-1}}{|z_n|^2}\right)^\alpha + \overline{c}R^2 \big\rvert^2=\big\lvert z_n^\alpha R^{-2\alpha} + \overline{c}R^2 \big\rvert^2=\big\lvert z_n^\alpha R^{-2(\alpha+1)} + \overline{c} \big\rvert^2=\big\lvert \overline{z_n}^\alpha  + c \big\rvert^2
\end{cases}
\end{equation}
So we must solve for $R$ where $R^{-\alpha-1}=1$ and we find that:
\begin{equation}
R \in \mathcal{U}_{\alpha+1}^{-1}=\{e^{\frac{-2 k \pi i}{\alpha+1}}:k \in [0,..,\alpha]\}
\end{equation} 
This corresponds to $\alpha+1$ symmetries when $\alpha \geq 2$ and $|\alpha|-1$ symmetries when $\alpha \leq -2$
