= Solution
For $F(t)=\prod_i(1+x_it)$ the coefficients are the <elementary symmetric functions>, $f_r=e_r(x)$. The specialization is evaluation in $x$ after the <symmetric-function involution>, because $\omega(h_r)=e_r$. Hence
$$
s_\lambda^F=\omega(s_\lambda)(x)=s_{\lambda'}(x),
\qquad
\boxed{s_\lambda^F=s_{\lambda'}(x).}
$$
One can also read this coefficient directly from the <dual Cauchy identity>, after exchanging its alphabets. For a general <symmetric function>, $u^F=(\omega u)(x)$; it is not simply $u(x)$.
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