Zoom out fractals? (A question about selfsimilarity)

It is well known that if we zoom in on the Mandelbrot set we get selfsimilarity.
So I wonder if $g$ is a fractal (in the complex plane) generated by a nonperiodic nonpolynomial entire function $f$

$g:= f(f(…))$

Is it possible that the fractal is infinite in size and that when we zoom out , we get selfsimilarity too ? Lets call such a fractal a ” zoom out fractal “.

As example Mandelbrot : $f(z)=z^2+1$, g$(z) [= f(f(…))]$ is the fractal.
The fractal has finite size (area or length ) since it diverges for $Re(z)>2$.
$f(z)$ is a nonperiodic nonpolynomial entire function. But when we zoom out we get no selfsimilarity. ( divergence is not considered valid as selfsimilar ) So Mandelbrot is NOT a zoom out fractal.

Does the existance of zoom out fractals require that the fractal is also a zoom in fractal ?

What is the formal way or term to express ‘ zoom out selfsimilarity ‘ or ‘ zoom out fractal ‘ , if any ?

Solutions Collecting From Web of "Zoom out fractals? (A question about selfsimilarity)"

A serious candidate for $f(z)$ is the following function : $\displaystyle f(z)=\lim_{\epsilon\to0}\int_\epsilon^{\infty} \dfrac{e^{zt}}{t^t} \, dt $.

The reason is that the entire function $f(z)$ is bounded for $\mathrm{Im}(z)^2 > \pi$

This was given here : Entire function $f(z)$ bounded for $\mathrm{Re}(z)^2 > 1$?

See also : How fast does the function $\displaystyle f(x)=\lim_{\epsilon\to0}\int_\epsilon^{\infty} \dfrac{e^{xt}}{t^t} \, dt $ grow?

It would be nice to see a plot.