Endomorphism ring spectrum of the Eilenberg-MacLane spectrum 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.
 A: No, $A$ is not an $H\Bbb Z$-algebra.
Suppose $R$ is an $H\Bbb Z$-algebra. Then the category of left $R$-modules is $H\Bbb Z$-linear: for any $R$-modules $M$ and $N$, the function spectrum $F_R(M,N)$ naturally has the structure of an $H\Bbb Z$-module. One reason for this is that $R$ is now an algebra object in the symmetric monoidal closed category of $H\Bbb Z$-modules, and the internal $R$-module function objects can be given weakly equivalent constructions there instead of in the category of $\Bbb S$-modules. In particular, the unit $\Bbb S \to F_R(M,M)$ of the endomorphism ring factors through $\Bbb S \to H\Bbb Z \to F_R(M,M)$ because the endomorphism ring is now an algebra in $H\Bbb Z$-modules.
However, if we take $R$ to be the Steenrod algebra spectrum and $M = H\Bbb F_p$ with the action it has by definition, there is an equivalence of the $R$-linear endomorphisms of $M$ with the $p$-completed sphere:
$$
F_A(H\Bbb F_p, H\Bbb F_p) \simeq \Bbb S^\wedge_p
$$
(This, for example, is what gives rise to the Adams spectral sequence.) In particular, the unit $\Bbb S \to \Bbb S^\wedge_p$ doesn't factor through $H\Bbb Z$ for any prime.
I think that many people find this pretty surprising when they first encounter it; I certainly did.
