Is there a Julia fractal that contains uncountable many copies of itself?

We know that the Mandelbrot fractal contains a countable number of copies of itself.

See :

Does the Mandelbrot fractal contain countably or uncountably many copies of itself?

Where that is explained.

Notice that polynomials have a finite amount of zero's and entire functions have a countable amount of zero's.

So I started to wonder :

Is there a Julia fractal that contains uncountable many copies of itself ?

And if so, can they be iterations of entire functions ?

What are typical examples ?

• Think about how e.g. you can't pack uncountably many disjoint "X-shapes" (for pretty much any interpretation of that phrase) in the plane. The same basic argument will apply to any set that's much more complicated than a line segment. Commented Nov 20, 2023 at 19:51
• @NoahSchweber If the size shape has a positive area on a surface or a positive volume in a space, you are probably correct.... But what if it does not ? Like some kind of sponge or so ?
– mick
Commented Nov 20, 2023 at 20:52
• If packing the shapes obligates that the "holes" between them have positive area (which I believe is the case NoahScheweber is asking you to consider) then from your comment I think you already agree with him, no? Commented Nov 20, 2023 at 20:57
• @RonKaminsky well it makes alot of sense. But I would not call it a proof yet ?
– mick
Commented Nov 20, 2023 at 21:33
• The Julia set of $z \to z^2 - 2$ is a line segment, which contains uncountably many line segments; however only countably-many are related to the dynamics, so this is more of a coincidence than anything structural. (Thanks for @NoahSchweber's hint.) Commented Nov 21, 2023 at 18:56