In unoriented cobordism there exist stable cohomology operations looking similar to Steenrod squares (they were used by Quillen to compute the unoriented cobordism ring with its formal group law defined by tensor product of bundles).

Since their properties are so similar to the properties of Steenrod squares, my questions are: do they become normally Steenrod squares in $N^*(X) \otimes_{N*} \mathbb{Z}_2 $? Do they correspond to $Sq^i \otimes id_{N^*} $ under the isomorphism $N^*(X) \simeq H^*(X) \otimes_{\mathbb{Z}_2} N^*$?

arethe power operations... So they definitely depend on the multiplicative structure, not just the additive structure. But you're right, I guess I don't know whether that map is anything more than E_2... It's likely not E_infty but I would be surprised if it wasn't H_infty (which is what we're asking). I dunno how I'd prove it... The analog for MU and BP is false I think, but with HF_2 I feel more optimistic. $\endgroup$3more comments