= Solution
Let $\widetilde W$ be the given regular irreducible family. Write $a=(x,y)$ and $a\wedge b=x_1y_2-y_1x_2$. Define the strongly integrated operator
$$
P=\frac1\pi\int_{\mathbb R^2}e^{-|a|^2/2}\widetilde W(a)\,da.
$$
The Gaussian is integrable and the operators are unitary, so the integral defines a <bounded operator> on each <vector>. The Weyl relation and $\widetilde W(a)^*=\widetilde W(-a)$ give $P^*=P$. For its square, the coefficient at $\widetilde W(c)$ is
$$
\frac1{\pi^2}\int e^{-(|a|^2+|c-a|^2)/2}e^{ia\wedge c}da
=\frac1\pi e^{-|c|^2/2},
$$
by the two-dimensional <Gaussian integral>. Thus $P^2=P$. The same Gaussian integration, with an inserted Weyl operator, gives
$$
\boxed{P\widetilde W(b)P=e^{-|b|^2/2}P.}
$$
This is the <Gaussian projection in a Weyl representation>.
It is nonzero. Otherwise all its conjugates would vanish; the Weyl relation makes their <matrix> coefficients the Fourier transforms of $e^{-|a|^2/2}(\widetilde W(a)\xi,\eta)$, with the dual variable $2b\wedge a$. Uniqueness of the <Fourier transform> of an integrable function would make each such function zero. <Strong continuity> then makes its value at zero vanish, contradicting $(\xi,\xi)>0$ for a nonzero <vector>.
Choose a unit <vector> $v\in PH$. Irreducibility makes the span of $\widetilde W(a)v$ dense. If $w\in PH$ is perpendicular to $v$, the boxed identity makes it perpendicular to this entire span, so $w=0$. Hence $P$ has <rank> one, and its identity gives $(\widetilde W(a)v,v)=e^{-|a|^2/2}$.
In the concrete representation of part (a), the Gaussian $v_0(t)=(2/\pi)^{1/4}e^{-t^2}$ has precisely this expectation. With the linear-in-first inner-product convention, both orbit families have <Gram matrix>
$$
(\widetilde W(a)v,\widetilde W(b)v)=e^{ia\wedge b}e^{-|a-b|^2/2}
=(W(a)v_0,W(b)v_0).
$$
Therefore the map $\sum c_jW(a_j)v_0\mapsto\sum c_j\widetilde W(a_j)v$ is well-defined and isometric. Its source and target spans are dense, so it extends to a <unitary operator> intertwiner. This proves the \b[<unitary equivalence>], the <Stone-von Neumann theorem> in the normalization of this question, rather than invoking that theorem as the proof.
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