= Solution
For the given <exchangeable> law $\mu$, put
$$
m_k=\mu(\eta\equiv1\text{ on }A)\quad\text{when }|A|=k,
$$
which is well-defined by the stated assumption. Integrate the <product self-duality of symmetric exclusion> against $\mu$, and use boundedness to interchange <expectation> and integration:
$$
\int\mathbb P^\eta(\eta_t\equiv1\text{ on }A)\,\mu(d\eta)
=\mathbb E^A\left[\mu(\eta\equiv1\text{ on }A_t)\right].
$$
The finite-particle process conserves its cardinality. Hence the right side is $\mathbb E^A m_{|A_t|}=m_{|A|}$. All occupation-product moments are therefore unchanged by evolution.
These moments determine the law, not merely its one-site marginals. For disjoint <finite sets> $B,C$, the <inclusion-exclusion principle> gives
$$
\mu(\eta\equiv1\text{ on }B,\ \eta\equiv0\text{ on }C)
=\sum_{D\subseteq C}(-1)^{|D|}\mu(\eta\equiv1\text{ on }B\cup D).
$$
Every term is unchanged, so every finite cylinder <probability> is unchanged. Such cylinder events generate the <product sigma-algebra> on $\{0,1\}^{\mathbb Z}$, and <probability> laws agreeing on them agree everywhere. Consequently
$$
\boxed{\mu P_t=\mu\quad(t\ge0).}
$$
Thus every <exchangeable> <probability> law is invariant for the <symmetric exclusion process>. This proof of <exchangeable invariant laws of symmetric exclusion> needs only the given cardinality dependence and duality; no representation theorem for <exchangeable> measures is required.
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