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Another answer already supplied an answer to the OP’s question, by pointing out that automorphisms of a group $G$ can act nontrivially on $H^2(G; F^\times)$. An alternate interpretation of the question is to find isomorphic twisted group algebras not related by an isomorphism of their grading groups. This is also easy: untwisted group algebras are special cases of twisted group algebras; if $F$ is algebraically closed, then both $F[\mathbb{Z}/4]$ and $F[(\mathbb{Z}/2)^2]$ are isomorphic to $F^{\times 4}$.