Let $X$ be a separated scheme of characteristic $p > 0$. I am going to assume that $F : X \to X$ (the absolute Frobenius) is a finite map. This condition is called being $F$-finite.

Question Suppose that $X$ has a dualizing complex $\omega_X^{\bullet}$. Is $F^! \omega_X^{\bullet}$ quasi-isomorphic to $\omega_X^{\bullet}$? In other words, is $R Hom_R^{\bullet}(F_* O_X, \omega_X^{\bullet})$ isomorphic to $F_* \omega_X^{\bullet}$ as $F_* O_X$-modules?

Special Cases If $X$ is of finite type over an $F$-finite field $k$, then this is true. One has the commutative diagram involving the absolute Frobenius on the base field and $X$, and one notices that $F^! k = Hom_k(F_* k, k)$ is abstractly isomorphic to $F_* k$ and the result follows.

In general of course, $F^! \omega_X^{\bullet}$ is still a dualizing complex, so it agrees with the original up to shift or twist by line bundles (and it's obvious that there are no shifts, so the only question is might one obtain a twist by a line bundle).

Remark According to Gabber, the assumption that $X$ has a dualizing complex is implied by $X$ being $F$-finite (see Remark 13.6 in O. Gabber: Notes on some t-structures, Geometric aspects of Dwork theory. Vol. I, II, Walter de Gruyter GmbH & Co. KG, Berlin, 2004, pp. 711–734). He provides a proof in the case that $X$ is affine.

Background I've talked to a few people about this question over the past couple years (in particular, Wenliang Zhang, who, iirc, first brought this question to my attention, as well as Gennady Lyubeznik, Joe Lipman, Shunsuke Takagi, and Manuel Blickle) but as far as I know, no one knows how to do it. Has anyone ever run into this question or a situation that looks similar?