So, following basically the same steps as everyone else, but with perhaps some slightly different naming. My personal cubic is:

$$f(z)=-441z^3-1278.9z^2-1234.8z-397.9$$

So I created two new functions $\phi(z)$ and $g_{a,b}(z)$ such that my polynomial is affinely conjugate to $g$ for some $a,b$ via $\phi$ for some $c,d$:

$$

\begin{array}{rcl}

g_{a,b}(z)&=&z^3-3a^2z+b\\

&\text{and}&\\

\phi(z)&=&cz+d

\end{array}

$$

I used Mackdaddica to solve:

$$(g_{a,b}\circ\phi )(z)=(\phi\circ g)(z)$$

for $a,b,c,$ and $d$:

```
f[z_] := -3979/10 - 12348/10 z - 12789/10 z^2 - 441 z^3
g[a_, b_][z_] := z^3 - 3 a^2*z + b
ϕ [z_] := c*z + d
eqs := CoefficientList[ϕ [f[z]] - g[a, b][ϕ [z]], z] == {0, 0,
0, 0}
eqs // InputForm;
sols = Solve[eqs, {a, b, c, d}]
(* Out:
{{b -> -d (-1 - 3 a^2 + d^2), c -> 0},
{a -> -((7 I)/10), b -> (7 I)/500, c -> -21 I, d -> -((203 I)/10)},
{a -> (7 I)/10, b -> (7 I)/500, c -> -21 I, d -> -((203 I)/10)},
{a -> -((7 I)/10), b -> -((7 I)/500), c -> 21 I, d -> (203 I)/10},
{a -> (7 I)/10, b -> -((7 I)/500), c -> 21 I, d -> (203 I)/10}}
*)
```

Like many others, I chose the solution where the values of $a,b$ were both positive. For me, this was where $a=\frac{7i}{10}$ and $b=\frac{7i}{500}$. Then I compared the Julia sets and, like any others, found my conjugated Julia set to be rotated, translated, and streched, but the same shape:

```
ff[z_] = g[a, b][z] /. sols[[3]]
Grid[{{JuliaSetPlot[ff[z], z, ImageSize -> {Automatic, 400}],
JuliaSetPlot[f[z], z, ImageSize -> 400,
PlotRange -> {{-1.05, -.88}, {-0.03, 0.03}}]}}, Spacings -> 5]
```

So now I had to test my two `parameterPic`

diagrams, one with $a$ fixed, and one with $b$ fixed

```
parameterPic["a", (7 I)/10, -1 - I, 1 + I, 0.005, Frame -> True,
PlotRangePadding -> None,
Epilog -> {Yellow, PointSize[Large], Point[{0, 7 /500}]}]
```

```
parameterPic["b", (7 I)/500, -1 - I, 1 + I, 0.005, Frame -> True,
PlotRangePadding -> None,
Epilog -> {Yellow, PointSize[Large], Point[{0, 7 /10}]}]
```