Is there ever a Kunneth isomorphism just for powers?

It's pretty rare for a multiplicative cohomology theory $$E$$ to have a Kunneth isomorphism $$E^\ast(X \times Y) \cong E^\ast(X) \otimes_{E^\ast(pt)} E^\ast(Y)$$ for all spaces $$X,Y$$. Are there any examples of cohomology theories $$E$$ which have a Kunneth isomorphism just for powers, but not for all products of spaces? That is,

Question: Are there examples of multiplicative cohomology theories $$E$$ such that for all spaces $$X$$ and all $$n \in \mathbb N$$, the canonical map $$E^\ast(X) \otimes_{E^\ast(pt)} \dots \otimes_{E^\ast(pt)} E^\ast(X) \to E^\ast(X\times \dots \times X)$$ is an isomorphism (where the tensor product and cartesian product are taken $$n$$ times), and yet there are spaces $$X,Y$$ such that the canonical map $$E^\ast(X) \otimes_{E^\ast(pt)} E^\ast(Y) \to E^\ast(X \times Y)$$ is not an isomorphism?

I'd be happy to understand this question in slightly different contexts, such as assuming $$X,Y$$ to be finite, or taking them to be spectra or pointed spaces rather than spaces, or assuming that $$E$$ has more structure.

As alluded to at the start, the condition that $$E$$ doesn't have a Kunneth isomorphism in general doesn't rule out too many $$E$$'s -- I think just $$H\mathbb F_p$$, $$K(n)$$, $$H \mathbb Q$$, and things constructed from them.

• Even Morava $K$-theory does not have a Kunneth isomorphism of the specified type. You would need to use completed tensor products, or work in homology rather than cohomology. Nov 28, 2019 at 23:00

The Künneth map is an isomorphism for trivial reasons if $$X$$ or $$Y$$ are empty. Otherwise, pick points $$x_0\in X,y_0\in Y$$; the idempotents $$p_x:x\mapsto x,y\mapsto y_0, p_y:x\mapsto y_0,y\mapsto y$$ induce the projections of $$E^*(X\sqcup Y)\cong E^*(X)\oplus E^*(Y)$$ onto its two summands. By naturality, the Künneth map $$E^*(X\sqcup Y)\otimes_{E^*(pt)} E^*(X\sqcup Y)\to E^*((X\sqcup Y)\times (X\sqcup Y))$$ preserves the decomposition of its source and target into the canonical four summands, and on each of these it is just the corresponding Künneth map. In particular, it is invertible if and only if the Künneth maps for $$(X,X),(Y,Y),(X,Y),(Y,X)$$ are all invertible. So your assumption on powers implies the Künneth map is always an isomorphism.