Consider a connected graph $\mathrm{G},$ locally finite (every vertex has finite neighbours), satisfying the following "circular condition": there exists a vertex $o$ such that for every pair of vertices $x$ and $y$ at the same distance from $o$ there exists a graph automorphism sending $x$ to $y$ (a bijection between vertices and edges that respects adjacency) while leaving $o$ fixed. If one think of planar graphs, any graph with the form of a spider web, though examples may arise in many dimensions.
Suppose a $h$ is a harmonic function on the vertices of $\mathrm{G}$ except at $o,$ where its value is specified to be one. This is to say, $h(o) = 1$ and for every other vertex $x \neq o,$ we have $h(x) = \dfrac{1}{d(x)} \sum\limits_{y \sim x} h(y),$ where $d(x)$ is the "degree" of $x,$ that is, the number of neighbours $x$ has.
Claim: the function $h$ satisfies that $h(x) = h(y)$ for every pair of nodes (vertices) $x$ and $y$ at the same distance from $o.$
The circular condition implies at once $d(x) = d(z)$ for all $x$ and $z$ at the same distance from $o.$ Any ideas and hints appreciated. Also, assuming the distance from $o$ to $z$ to be $n,$ I have tried to expand $h(z) = \dfrac{I + J + K}{d(x)},$ where $I$ is the sum of $h(t)$ for $t$ a neighbour of $z$ and at distance $n - 1,$ $J$ the sum for vertices at distance $n$ and $K$ the sum for distance $n + 1$ (there can't be other type neighbours of $z$). But the problem is that the equations become really messy and I suppose there must be simpler methods than brute force.
Think of an infinite binary tree with $o$ being its root. Then assign the values for $h(x)$ as follows.
Left branch: $h(x)={1\over2}$ for the vertex adjacent to the root, then $1\over4$ for its children, then $1\over8$ for the next level, and so on.
Right branch: $h(x)=1$ for all $x$.
...
So no, the claim does not hold.
Come to think of it, why would you call it a voltage function? Supposedly because it represents the voltage values in an electric network shaped according to our graph, if you touch it with electrodes at certain points (those where we specify $h(x)$). But wait, there is only one such point ($o$), and we can't have any currents in a setup with only one wire attached, can we? Why, we can, only the other electrode(s) must be removed to infinity. Can we have many of those? Sure, why not. Can they have different voltages?
Well, that's just what I did.