Conditions imposed on graph automorphism leave the "voltage" on graph nodes identical? We call $\pi$ a graph automorphism on a graph $(V,E)$ if $\pi:V\to V$ is a bijection such that $xy\in E\iff \pi(x)\pi(y)$ is an edge, and $\phi(xy)=\phi(\pi(x)\pi(y))$, where $\phi$ is the conductance.  
I want to show that, given two distinct points $a,b$ we have that $f(x)=f(z)$, if it holds that $\pi(a)=a$, $\pi(b)=b$, $\pi(x)=z$. 
Here, $f$ is some arbitrary harmonic function (in the sense of stochastic processes) at every node $x\in V\backslash\{a,b\}$, that is $$f(x)=\sum_{xy\in E}\frac{\phi(xy)}{\phi(x)}f(y)$$
Here, the conductance on a vertex/node $\phi(x)$ is the sum of all the conductances on the adjacent edges.
I am not sure how to visualise a graph automorphism that satisfies the stated conditions, other than trivial ones. So I continued without trying to use that intuition, but got stuck all the same.  
I tried writing
$$f(x)=\sum_{xy\in E}\frac{\phi(xy)}{\phi(x)}f(y)=\sum_{xy\in E}\frac{\phi(\pi(x)\pi(y))}{\phi(x)}f(y)=\sum_{xy\in E}\frac{\phi(z\pi(y))}{\phi(x)}f(y)$$
but I don't see how the given conditions can be used to further modify this expression into the desired form (which would be $\sum_{zk\in E}\frac{\phi(zk)}{\phi(z)}f(k)$).
In particular, I think I'm missing how to include the relevant conditions imposed on $\pi(a)$ and $\pi(b)$. Could anyone give some suggestions as to how to proceed?
 A: I'll asume that $(V,E,\phi)$ is a connected, finite, weighted graph and that $f$ is harmonic on $V\setminus \{a,b\}$ (if the graph is not connected, then any such $f$ will be constant on all of its components save for the common component of $a$ and $b$). Let $(X^{x_0}_n)_{n\in\mathbb{N}}$ denote the associated simple random walk started at $x_0$. 
That is, 
$$
\mathbb{P}(X^{x_0}_{n+1}=y|X^{x_0}_n=w)=\begin{cases} \frac{\phi(wy)}{\phi(w)} & w\not\in \{a,b\}, wy\in E \\
1 & w=y\in \{a,b\} \\
0 & else\end{cases}
$$
and $X_1^{x_0}=x_0$.
Then, by assumption, $(\pi(X_n^{x}))_{n\in \mathbb{N}}$ has the distribution of $(X_n^{z})_{n\in \mathbb{N}}$, and hence $(f(\pi(X_n^{x})))_{n\in \mathbb{N}}$ and $f(X^x_n)_{n\in \mathbb{N}}$ are both bounded martingales (since $f$ is harmonic). Let $\tau$ denote the first hitting time of $\{a,b\}$. Then, applying the optional sampling theorem, we get 
\begin{align}
f(x)=\mathbb{E}f(X_1^x)=\mathbb{E} f(X_{\tau}^x)&= f(a)\mathbb{P}(X^x_{\tau}=a)+f(b)\mathbb{P}(X^x_{\tau}=b)\\
&= f(a)\mathbb{P}(\pi(X^x_{\tau})=a)+f(b)\mathbb{P}(\pi(X^x_{\tau})=b) \\
&= f(a)\mathbb{P}(X^{z}_{\tau}=a)+f(b)\mathbb{P}(X^z_{\tau}=b) \\
&= \mathbb{E} f(X^z_{\tau}) \\
&= f(z)
\end{align}
