Let G be a finite group. Is there always a finite group H such that G≅Aut(H)?

This is false in general. Indeed we will prove that if m is an odd positive integer then doesn’t exist a finite group H such that ℤm≅Aut(H). Indeed if such a group exists then Aut(H) is cyclic and not trivial, by the last post at link https://marcodamele.blog/2024/09/05/let-g-be-a-group-such-that-autg-is-cyclic-and-not-trivial-then-autg-is-finite-and-his-order-is-even/ we would have that Aut(H) is even which contradict the fact that |Aut(H)|=|ℤm|=1 mod 2. ■