How fast a hexagon smooths, and why self-loops matter

The distance shrinks by at most the largest eigenvalue below $1$ in absolute value per layer. Without self-loops every degree is $2$.

The eigenvalues other than $1$ are $\tfrac23$, $\tfrac23$, $0$, $0$ and $-\tfrac13$, so $|\lambda_2|=\tfrac23$ and the distance after $k$ layers is at most $(\tfrac23)^k$ times its start. $(\tfrac23)^{11}=0.011561$ and $(\tfrac23)^{12}=0.007707$, so $12$ layers suffice for any starting signal, and a signal with a component along the slowest mode needs that many.

Without self-loops the operator is $\mat A/2$, and its eigenvalues are $\cos\theta$ for $\theta=2\pi j/6$: $1$, $\tfrac12$, $\tfrac12$, $-\tfrac12$, $-\tfrac12$ and $-1$. The eigenvalue $-1$ belongs to the alternating signal: each node’s two neighbours have the opposite sign, so the average is exactly the negative. The signal flips sign at every layer and never converges. The limit of I.14.T3 does not exist.

The hexagon is bipartite, since its nodes split into alternate halves with every edge between them, and every bipartite graph has this eigenvalue. The self-loop mixes each node with itself. On a graph where every degree is $r$ it maps each eigenvalue $\lambda$ of the plain operator to $(r\lambda+1)/(r+1)$, so $-1$ becomes $(1-r)/(r+1)$, which is $-\tfrac13$ here. On any connected graph it rules out $-1$.