If $ |f(z^{2})|\leq|f(z)| $ and $ f $ is entire, then is $ f $ constant?

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Assume $f:\mathbb{C}\to \mathbb{C} $ entire function, and assume $ f $ satisfies $ |f(z^{2})|\leq|f(z)| $

How can I prove that $ f $ is a constant?

I have tried something but not sure about my way. I'll share my attempt any way:

Let $ \overline{D(0,R^2)} $ be a closed disk with $R^2>1 $. Then by the maximum modulus theorem, $ f $ accepts its maximum on $ C(0,R^2) $, i.e. on some point $ R^{2}e^{i\theta} $. But then we have $ |f(R^{2}e^{i\theta})|\leq f(Re^{i\frac{\theta}{2}})| $ So $ f $ actually assumes its maximum in an inner point of $ \overline{D(0,R^{2})} $, and thus $ f $ is constant in the closed disk, and by the uniqueness theorem, in the whole plane.

I feel like something's wrong with the proof. Any clarifications would be helpful.