Consider the endomorphism ring spectrum $R = \mathrm{End}_S(H\mathbb{F}_p)$ of the mod $p$ Eilenberg-MacLane spectrum $H\mathbb{F}_p$. The homotopy groups of $R$ are the Steenrod algebra $A^*$ with reversed grading:
$$\pi_n R = [\Sigma^n H\mathbb{F}_p, H\mathbb{F}_p] = A^{-n}.$$
This spectrum $R$ is an associative $S$-algebra (or $A_{\infty}$ ring spectrum). Moreover, $R$ is an $H\mathbb{F}_p$-module spectrum, say, using the $H\mathbb{F}_p$-module structure on the target $H\mathbb{F}_p$. In particular, $R$ is an $H\mathbb{Z}$-module spectrum. However, $R$ is known *not* to be an $H\mathbb{F}_p$-algebra spectrum.

Question. Is $R = \mathrm{End}_S(H\mathbb{F}_p)$ an $H\mathbb{Z}$-algebra spectrum?

My hunch is that the answer is no, but I couldn't find that statement in the literature. Perhaps it can be shown using an invariant of structured ring spectra, some flavor of $THH$. Or perhaps a dg-algebra over $\mathbb{Z}$ doesn't have enough room to encode the homotopical structure of the Steenrod algebra [1].

**Remark**. For the sake of definiteness, feel free to pick a model of spectra such as $S$-modules or symmetric spectra. The question is meant to be about the underlying symmetric monoidal $\infty$-category. In light of [2], working with your favorite model of spectra should be fine.

[1] *Shipley, Brooke*, **$H\Bbb Z$-algebra spectra are differential graded algebras**, Am. J. Math. 129, No. 2, 351-379 (2007). ZBL1120.55007.

[2] *Mandell, M.A.; May, J.P.; Schwede, S.; Shipley, B.*, **Model categories of diagram spectra**, Proc. Lond. Math. Soc., III. Ser. 82, No.2, 441-512 (2001). ZBL1017.55004.