I would like to find an entire, non-constant (complex) function $f$ with $f(cz) = f(z)$ $\forall z \in \mathbb{C}$ for some $c \in \mathbb{C}$.
This is pretty straightforward if $|c| \neq 1$, but I am going to establish a few properties of $f$ first:
Because $f$ is entire, we know that it is continuous on the unit disk $B_0(1)$ and thus bounded on $B_0(1)$. Because $f(cz) = f(z)$ for any $z$, we can repeatedly plug in $cz$ in to the right side and get $f(c^k z) = f(z)$ $\forall k \in \mathbb{N}$.
Suppose $|c| > 1$ and let $z_0 \in \mathbb{C}$. Then $|z_0| < |c|^k$ for $k \geq N$ because $|c|^k$ diverges. This can be rearranged to $|\frac{z_0}{c^k}| < 1$, so $\frac{z_0}{c^k} \in B_1(0)$ and with the second property $f(z_0) = f(c^k \cdot \frac{z_0}{c^k}) = f(\frac{z_0}{c^k})$. This means that for any $z_0$, $f(z_0)$ corresponds a value on the unit disk (whose values are bounded) and thus $f$ is bounded. Since $f$ is entire and bounded, Liouville's theorem tells us that $f$ is constant. So for any $c$ with $|c| > 1$, there is no non-constant function $f$ with this property.
The case $|c| < 1$ follows analogously with $|z_0| < |\frac{1}{c}|^k$ for $k \geq N$.
Now suppose $|c| = 1$. Write $c$ in polar coordinates: $c = e^{\varphi i}$ with $\varphi \in [0,2\pi)$. If $\varphi \in \pi \mathbb{Q} := \{\pi \cdot q | q \in \mathbb{Q}\}$, we can again write $c = e^{\frac{p}{q}\pi i}$ with $p,q \in \mathbb{N}$ and $\text{gcd}(p,q) = 1$. One can now take $f(z) = z^{2q} \in \mathcal{O}(\mathbb{C})$ (meaning $f$ is entire) and with this
$f(cz) = (cz)^{2q} = (e^{\frac{p}{q}\pi i})^{2q} \cdot z^{2q} = e^{2p\pi i} \cdot z^{2q} = z^{2q} = f(z)$
so $f(z) = z^{2q}$ is of the desired form.
The case that gets me stumped is when $\varphi \not\in \pi \mathbb{Q}$ or if there is a much easier way that I am overlooking. In any way, I am interested if there are any more such functions with the properties I want (but especially for the case that is still missing). For orientation, I am an undergraduate that has just taken his first complex analysis course this year. Any help is appreciated!