Let $f:\mathbb{C} \rightarrow \mathbb{C}$ be an entire function and $\xi=e^{\frac{2\pi i}{n}}$ for some $n\in \mathbb{N}$. Suppose that $f(\xi z)=f(z)$ for all $z\in \mathbb{C}$ and consider the function $$g(z)=\frac{1}{n}\sum_{k=0}^{n-1} f \left(\xi^{k}\sqrt[n]{z}\right)$$ where $\sqrt[n]{z}$ is the $n$th root with argument between $0$ and $2π/n$. Then $g$ is an entire function.
Remark: I think the problem is reduced to show that the function $\sqrt[n]{z}$ with the argument between $0$ and $2π/n$ is an entire function.
Expand $f(z)=\sum_m c_m z^m $ in it's Taylor series. Consider the cyclic group $(\Xi =\{ \xi ^k\}_{k=1}^n)$ of integer powers of the primitive root $ \xi= e^{2\pi i /n}$ that satisfies $\xi^n=1$. Take the average of the values of $ f( \xi^k z)$ as $\xi^k$ ranges over this finite cyclic group, by averaging termwise across the Taylor series. An easy exercise confirms that only certain monomial powers in the Taylor expansion of $f(z)$ survive this averaging, namely those powers that are $ z^n, z^{2n} , etc$; and averaging leaves these terms invariant. Thus $g(z)= \sum_{r} c_{mr} z^{nr}$. This new function is entire because it is the sum of finitely many modified versions of $f$.