Is there a nontrivial analytic function which takes the same values on disjoint circles? 
Given two disjoint circles $C_1$ and $C_2$ in the complex plane, is there a nontrivial analytic function $f(z)$ whose values on $C_1$ are the same as its values on $C_2$ up to rotation and dilation?

What little intuition I have about this suggests not. If we had a function $f = u+i~v$ and think of the harmonic functions $u,v$ as potential distributions, we could consider the special case of two nearly osculating circles of very different size. Informally (and far too vaguely) the mutual patterns of influence of the two circles seem qualitatively different, so it seems unreasonable to suppose the functions could be the same. This argument does not work for circles of the same size.       
As a guess we would have that $f(z)=u(x,y)+i~v(x,y) = f(w(z))$ for some Mobius transformation $w(z)$ and maybe the Cauchy-Riemann equations cannot be satisfied? I was hoping for an obvious geometric argument but maybe it's not possible? 
Edit: It seems that this may be possible for same-radius circles (I didn't want to rule these out) and I would upvote/accept an answer along these lines, but would add 100 points to an answer that proves it's impossible [re-edit: and kal v'chomer possible] for different-radius circles. 
 A: Since an entire function is completely determined by its value on a circle, your requirement is in fact very, very strong.
Let $\tau : \Bbb C \to \Bbb C$ be a composition of a translation a rotation and a scaling (which means a map $z \mapsto az+b$ for some $a \in \Bbb C^*$ and $b \in \Bbb C$).
If $f \circ \tau = f$ on some circle $C$, then we must have $f \circ \tau = f$ on all of $\Bbb C$.
If $a \neq 1$, then $f$ has a fixpoint $z_0 = b/(1-a)$, and 
$f$ has Taylor series around $z_0$, $f(z) = \sum c_k (z-z_0)^k$.
The equation now says $f(z) = f(\tau(z)) = \sum c_k (\tau(z)-z_0)^k = \sum c_k (a(z-z_0))^k  $, and so $c_k = a^kc_k$ forall $k$.
From there we get that if $a$ is not an $n$th root of unity for some integer $n$ then $f$ has to be constant, and if it is an $n$th root of unity, then $f$ is a function of $(z-z_0)^n$. For exemple, $f(z) = z^2$ is symmetric around $0$ so you can find two circles with the same image, and so on.
If $a=1$ then $\tau$ is a translation, and the equation simply says that $f$ is $b$-periodic. Now it is a classic result that the $b$-periodic entire functions are exactly the functions of $\exp(2i\pi z/b)$.
A: Here's an even simpler proof that this is impossible for nonconstant entire functions $f$ when the circles $C_1$ and $C_2$ are of different sizes:
Suppose there is a rotation and dilation $\tau : \mathbb{C} \to \mathbb{C}$, $\tau(x) = ax+c$, $\tau(C_1) = C_2$, such that $(f \circ \tau)|_{C_1} = f|_{C_1}$.
Since by the principle of permanence an analytic function is entirely determined by its values on $C_1$, we get $f \circ \tau = f$, hence $f \circ \tau^n = f$ for all $n \in \mathbb{Z}$.
Now choose any $x \in C_1$ that is not a fixed point of $\tau$, and let $A = \{\tau^n(x) : n \in \mathbb{Z}\}$. We know that $f$ is constant on $A$. If $A$ has an accumulation point, then $f$ is constant on $\mathbb{C}$, again by the principle of permanence. The only way for $A$ not to have an accumulation point is if $a$ is a root of unity. This of course implies that $C_1$ and $C_2$ have the same size.
