How big can $U\subset\mathbb{C}$ be if there exists a non-constant holomorphic $f\colon U\to\mathbb{C}$ with $2f(2z)=f(z)+f(z+1)?$

Although this answer does not fully address OP's question, I hope it provides some useful information on it.

Here, we will assume:

• $X \subseteq \mathbb{C}$ is a set containing an open neighborhood $U$ of $[0, 2]$.

• For each $z \in X$, both $\frac{z}{2}$ and $\frac{z}{2}+1$ are elements of $X$. (For instance, this holds when $X$ is a convex set containing $U$.)

• $f : X \to \mathbb{C}$ is a function that satisfies $$ f(z) = \frac{1}{2}\left( f\left(\frac{z}{2}\right) + f\left(\frac{z}{2}+1\right) \right) \tag{1} $$ for any $z \in X$.

• $f$ is continuous on $U$.

Then we claim that $f$ is constant. Indeed, it is straightforward to verify that

$$ f(z) = \sum_{k=0}^{2^n - 1} f\left(\frac{z}{2^n} + \frac{2k}{2^{n}} \right) \frac{1}{2^n} \tag{2} $$

holds for all $n \geq 1$ and $z \in U$. So by the continuity assumption, as $n\to\infty$ we have

$$ f(z) = \int_{0}^{1} f(2x) \, \mathrm{d}x, $$

which is independent of $z$. Therefore any such $f$ must be constant.

Here are some quick follow-up questions:

1. What can we say about $f$ when the equation $\text{(1)}$ is required to hold only when all of $z$, $\frac{z}{2}$, and $\frac{z}{2}+1$ are simultaneously in $X$ (as in OP's original formulation)?

2. The identity $\text{(2)}$ seems to suggest that there might exist a holomorphic function on $\mathbb{C}\setminus[0,2]$. Is it indeed possible to pursue in this direction?

Note that $\cot\frac{\pi (t+1)}{2}=-\tan(\frac{\pi t}{2})$ hence $\cot(\frac{\pi t}{2})+\cot\frac{\pi (t+1)}{2}=2\frac{\cos (\pi t)}{\sin (\pi t)}=2 \cot (\pi t)$

hence $f(t)=\cot(\frac{\pi t}{2})$ satisfies $f(t)+f(t+1)=2f(2t)$ and is analytic on $\mathbb C-2\mathbb Z$