Is boolean formula isomorphism NP-complete?
Your problem is $\mathrm{co}$-$\mathrm{NP}$-hard so it may also be $\mathrm{NP}$-hard but very unlikely to be $\mathrm{NP}$-complete (i.e. to be in $\mathrm{NP}$).

We reduce from TAUT which is $\mathrm{co}$-$\mathrm{NP}$-hard. Given a TAUT instance $\varphi$ with variables $x_1,\dots x_n$, output $&lt;\varphi, ((x_1 \lor \lnot x_1)\land\dots\land (x_n \lor \lnot x_n))&gt;$. These two formulae are isomorphism iff. $\varphi$ is a tautology.