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Counterexample. Let $E_1=\{e_0,e_1\}$ and $E_2=\{e_0,e_2\}$ where $$e_0=\{0,1\}\cup\{2,4,6,8,\dots\},$$ $$e_1=\{0\}\cup\{3,5,7,9,\dots\},$$ $$e_2=\{0,1\}\cup\{3,5,7,9,\dots\}.$$ Then $(\omega,E_1)\not\cong(\omega,E_2)$, but $(E_1,V_1^*)\cong(E_2,V_2^*)$ since $V_1^*=\{\{e_0\},\{e_1\},\{e_0,e_1\}\}$ and $V_2^*=\{\{e_0\},\{e_2\},\{e_0,e_2\}\}.$