holomorphic functions, roots of unity and harmonic numbers

Question

If $f$ is a non-constant holomorphic function such that, for all $z \in \mathbb{C}$, exists a $c \in \mathbb{C}$ where $f(cz) = f(z),$ then $c$ must be a $n$-th root of unity, or there exists some counterexample?

Furthermore, there exists $f$ holomorphic function and $n \in \mathbb{N}$ such that $$f(z\mathcal{H}(n)) = f(z), \forall z \in \mathbb{C}?$$

In this case, $\mathcal{H}(n)$ denotes the $n$-th harmonic number: $$ \mathcal{H}(n) = \sum\limits_{k = 1}^n \frac{1}{k} $$

Answer 1

Regarding your first question: If you allow $c$ to depend on $z$, there is very little one can say about $c$.

Consider $f(z) = \sin z$. Given $z \ne 0$, set $c = \frac{z + 2 \pi}{z}$. For $z = 0$, set $c = 2$. Then $f(cz) = f(z)$ for all $z$.

Assume now that $\exists c \, \forall z \, f(z) = f(cz) = f(c^{-1}z)$ and $f$ is not constant and entire. Then $c$ must be a root of unity.

Clearly $|c| = 1$, since otherwise we have the sequence $c^k$ or $c^{-k}$ converging to $0$ on which $f$ is constant, implying that $f$ is constant. If $c$ is not a root of unity, then $\{c^n : n \ge 0\}$ is dense on the unit circle and therefore $f(z) = f(1)$ on the unit circle, again implying that $f$ is constant.

Answer 2

For the second part: replace $z$ by $\frac z {H(n)}$ to get $f(z)=f(\frac z {H(n)})$. By iteration $f(z)=f(\frac z {H(n)^{k}})$ for any $k \geq 1$. Letting $k \to \infty$ we get $f(z)=f(0)$ for all $z$ provided $n >1$. For $n=1$ there is no hypothesis, so $f$ can be any entire function.