Consider a simple random walk on a undirected, connected graph. This is the random walk which, at every time step, moves to a random neighbor, with all neighbors being equally likely. Lets assume every node has a self-loop to avoid issues associated with periodicity. Then, the following (surprising to me) fact is true: under the stationary distribution, every edge is traversed with the same probability.
I'm trying to get some intuition for why this holds, and I am wondering whether someone can provide an explanation for this phenomenon.
Naturally, I know the standard proof which observes that $\pi_i = d_{i}/\sum_k d_k$ satisfies the equations for the stationary distribution, and so $\pi_i p_{ij} = 1/\sum_k d_k$ whenever the edge $(i,j)$ is present in the graph. This proof, however, feels more like a lucky calculation to me than an explanation.
Note that the fact in boldface is false for directed graphs. Its also false for the corresponding two-step chain, i.e. the Markov chain with probability transition matrix $P^2$, where $P$ is the transition matrix of the simple random walk.