The inverse Galois problem asks whether every finite group appears as the Galois group of a Galois extension of the rational numbers.


Is anything known about the anologous problem, where the extensions are not required to be Galois? In other words, for a finite group $G$, does there exist a finite field extension $K$ of $\mathbb{Q}$ such that $\mathrm{Aut}(K/\mathbb{Q})=G$?


Is this suspected to be as difficult as the inverse Galois problem or easier?



Inverse Galois problem for non-Galois extensions
The answer to this is positive. The first correct proof, it seems, was given in


Michael D. Fried. A note on automorphism groups of algebraic number
fields. Proc. Amer. Math. Soc., 80(3):386–388, 1980.


For a generalization to Hilbertian fields and some history see for example


F. Legrand and E. Paran. Automorphism groups over Hilbertian fields. Journal of Algebra
Volume 503, 2018. [journal website](https://doi.org/10.1016/j.jalgebra.2017.12.041)


