Is there an analytic function with $f(z)=f(e^{iz})$? 
Does there exist a non-constant analytic function $f$ satisfying
  $f(z)=f(e^{iz})$ ?

I don't know where to start.
 A: We are looking for $f$ with  $f=f\circ g$ where $g(z)= e^{iz}$. We notice that $g$ happens to have a fixed point $w\approx0.576+0.375 i$. As $|g'(w)|=|g(w)|=|w|<1$ it is an attracting fixpoint. Consequently, there exists a sequence $z_0,z_1,\ldots$ of points with $z_{n+1}=g(z_n)$ converging to $w$ (we just need to pick $z_0$ sufficiently close to but different from $w$). We conclude that $f$ is constant on the set $\{z_0,z_1,\ldots\}$ with accumulation point $w$ and so by the identity theorem $f$ is constant.
Note that this proof does not require $f$ to be entire, it suffices that its domain $D$ is connected and contains a neighbourhood of $w$ (also, we require $f(z)=f(g(z))$ only for $z\in D$ with $g(z)\in D$).
A: We have, for any entire function $g(z)$, the following formula for the $n$th derivative of $g(e^{iz})$
$$
\frac{d^ng(e^{iz})}{dz^n}=\sum_{k=1}^na_{k,n}e^{ikz}g^{(k)}(e^{iz})\qquad(*)
$$
for some constants $a_{k,n}$. A closer examination shows that
$$
a_{k,n}=i^n b_{k,n}
$$ 
for some positive integers $b_{k,n}$ such that $b_{n,n}=1$. This is easy to see by induction on $n$.
The function $h(z)=e^{iz}$ has at least one fixed point $z_0$. I "found" one such fixed point $z_0\approx0.576413 + 0.374699 i$ trivially by iterating $h$
(we should be able to prove this conclusively - leaving this step open for now).
Let's now turn our attention to an entire function $f$ satisfying the equation $f(z)=f(e^{iz})$ for all $z$. By formula $(*)$ we have for the $n$th derivative
$$
f^{(n)}(z_0)=f^{(n)}(e^{iz_0})=\sum_{k=1}^ni^nb_{k,n}e^{ikz_0}f^{(k)}(e^{iz_0})=
\sum_{k=1}^ni^nb_{k,n}z_0^kf^{(k)}(z_0).\qquad(**)
$$
Because $|z_0|<1$ and $b_{n,n}=1$ we have that the coefficients of $f^{(n)}(z_0)$ on opposite sides of $(**)$ are not equal. Therefore the system of linear equations $(**)$, by induction on $n$, imply that $f^{(n)}(z_0)=0$ for all positive integers $n$. 
Therefore the Taylor series of $f(z)$ centered at $z_0$ reduces to a constant. Because $f(z)$ is entire, this implies that $f$ must be a constant.
A: This is a very partial "answer", but it may give people some ideas. I'm trying to answer the question:

Does there exist a non-constant entire (analytic everywhere) function $f$ satisfying $f(z)=f(e^{iz})$?

Suppose that $f$ is a non-constant entire function satisfying $f(z)=f(e^{iz})$. 
Then, by definition there must exist constants


*

*$a_0$

*$a_1,a_2,...$, not all $0$


such that $f(z)=\sum_{n=0}^\infty a_n z^n .$ So,
$$
f(e^{iz})
= \sum_{m=0}^\infty a_m e^{imz} 
=\sum_{m=0}^\infty a_m \sum_{k=0}^\infty \frac{(imz)^k}{k!}.
$$
Changing the order of summation,
$$f(e^{iz})=\sum_{n=0}^\infty \sum_{m=0}^\infty a_m\frac{(imz)^n}{n!} = \sum_{n=0}^\infty z^n\sum_{m=0}^\infty a_m\frac{(im)^n}{n!}.$$
Using the assumption that $f(z)=f(e^{iz})$, we equate coefficients to see 

$$ a_n = \sum_{m=0}^\infty a_m\frac{(im)^n}{n!} 
 = \frac{i^n}{n!}\sum_{m=0}^\infty a_m m^n, \forall n\ge 0 .$$

Is it possible for this recurrence to be satisfied under the above assumptions?
