I want to know if there exists a non-trivial ($h \neq 0$) continuous function $h:S^1 \rightarrow \mathbb{C}$ such that for all $z \in S^1$, $$ h(z) = h(\sqrt z) + h(- \sqrt z). $$
My suspicion is that this is not true. I tried starting by noting that $$ h(1) = h(1) + h(-1) \implies h(-1) = 0, $$ thus we can proceed in this fashion and show that $h(i)+h(-i)=0$, we can keep doing this, however this seems like a dead end to me.
Consider $f(t)=\sum_{k \ge 0}\frac{e^{2^{k+1}t \pi i}}{2^k}$.
Clearly, $f$ is continuous on $\mathbb R$ and periodic with period $1$.
Since $e^{2t \pi i}+e^{2(t+\frac{1}{2})\pi i}=0$ while $e^{2^{k+1}t \pi i}=e^{2^{k+1}(t+\frac{1}{2}) \pi i}, k \ge 1$ we have
$f(t+\frac{1}{2})+f(t)=2\sum_{k \ge 1}\frac{e^{2^{k+1}t \pi i}}{2^k}=\sum_{k \ge 0}\frac{e^{2^{k+1}(2t) \pi i}}{2^k} =f(2t)$
But then if $z=e^{2\pi it}, t \in \mathbb R$ defining $h(z)=f(t)$, so $h(-z)=f(t+\frac{1}{2}), h(z^2)=f(2t)$, we have that $h$ is well defined on the circle (because $f$ periodic of period $1$), continuous and $h(z^2)=h(z)+h(-z)$ so the required property is satisfied.
One can construct many such examples using other sets of the type $(2m+1)2^k, k=0,1...$ and $m$ fixed (and unions of such with appropriate coefficients so we still have absolute convergence) and defining appropriate Fourier coefficients as above so they satisfy $a_{(2m+1)2^k}=2a_{(2m+1)2^{k+1}}$, so essentially $a_{(2m+1)2^k}=\frac{c_m}{2^k}, k \ge 0$; the example above is simplest where we just base our functions on $2^k$ (so $m=0$) and take $a_1=1$