= Solution
Let the common <vertex> degree be $r>0$. For each unordered <edge> place an independent <Poisson process> of rate $1/r$, and at a mark interchange the two endpoint occupations. If exactly one endpoint is occupied, this moves its particle across that <edge> at rate $1/r$; if both or neither are occupied, the occupied set stays unchanged. Thus each particle proposes jumps at total rate one, uniformly among its neighbours, with occupied-target proposals suppressed. This is the stated <symmetric simple exclusion process>. An edgeless <graph> has the identity dynamics and both conclusions below hold trivially.
In the <stirring representation of symmetric exclusion>, put a distinct label at every <vertex> and interchange labels at every <edge> mark, even when the occupations agree. There are finitely many marks on every finite time interval. Their chronological composition is a random <permutation> $\pi_t$ of the <vertex> set, and simultaneously for every initial set,
$$
\eta_t^A=\pi_t(A).
$$
Each <edge> swap is its own inverse. Reflecting the <edge>-mark times in $[0,t]$ preserves their joint <Poisson process> law and reverses the order of the swaps. Hence $\pi_t^{-1}$ and $\pi_t$ have the same distribution. Pathwise,
$$
\pi_t(A)\supseteq B\quad\Longleftrightarrow\quad\pi_t^{-1}(B)\subseteq A.
$$
Taking <probabilities> and then using the inverse-<permutation> distribution proves the <product self-duality of symmetric exclusion> in set form:
$$
\boxed{\mathbb P(\eta_t^A\supseteq B)=\mathbb P(\eta_t^B\subseteq A)}.
$$
No assumption about disjointness or equal cardinality of $A$ and $B$ is required; impossible inclusions have <probability> zero on both sides.
For invariance, independently choose the initial occupied set $A$ under the <product measure> $\mu_\rho$. The occupations are independent identically distributed <Bernoulli distribution> variables, so permuting their sites by any deterministic <permutation> preserves their joint law. Conditional on the independent stirring <permutation> $\pi_t$, the output therefore still has law $\mu_\rho$. Averaging proves the <Bernoulli invariant laws of finite symmetric exclusion>:
$$
\boxed{\mu_\rho P_t=\mu_\rho\qquad(0\le\rho\le1)}.
$$
There is also a direct check from the duality. For every $B\subseteq V$,
$$
\int\mathbb P(\eta_t^A\supseteq B)\,\mu_\rho(dA)
=\mathbb E\left[\mu_\rho\{A:A\supseteq\eta_t^B\}\right]
=\mathbb E\rho^{|\eta_t^B|}=\rho^{|B|},
$$
because the <exclusion process> conserves particle number. These are exactly the joint occupation moments of $\mu_\rho$; on a finite set they determine the law by the <inclusion-exclusion principle>. The endpoint densities give the absorbing empty and fully occupied configurations, and the empty-set moment is one also at $\rho=0$.
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