I would like an example of the following situation, or a proof that no such example exists.


$\textbf{Situation}$: Two connective (EDIT: I'm fine with dropping this condition) spectra $X$ and $Y$ such that the spaces $\Omega^\infty\Sigma^nX$ and $\Omega^\infty\Sigma^nY$ are equivalent (as spaces) for all $n$, but as spectra, $X$ and $Y$ are $\textit{not}$ equivalent. It was pointed out to me by Maxime that another way to phrase things is that $X$ and $Y$ are two (grouplike) $E\_\infty$ spaces that are equivalent as $E\_n$-spaces for all finite $n$ but not as $E\_\infty$-spaces.


I think (as Charles pointed out in the comments, this is wrong) it would be sufficient to find an example of the following: An infinite loop space $A$ (with terms $A\_0=\Omega A\_1,A\_1,...$ and structure maps $a\_i:A\_i\rightarrow \Omega A\_{i+1}$ such that $a\_0$ is the identity) and an equivalence $f:A\_0\rightarrow A\_0$ (of spaces) such that for any (infinite loop) equivalence $\phi:A\rightarrow A$ with 0th component $\phi\_0:A\_0\rightarrow A\_0$, the composite $\phi\_0\circ f$ is not homotopic to the component of any infinite loop map. Because then we can consider the infinite loop space $B$ with the same terms $B\_i=A\_i$ and the same structure maps $b\_i=a\_i$ for $i>0$ but with the $b\_0$ changed to $f$. The condition on $f$ implies that there is no way to fill in the 0th component of any attempted equivalence between $A$ and $B$.


The problem is that due to my insanely low supply of brain cells the only infinite loop spaces I can really come to grips with are Eilenberg-Maclane, and among Eilenberg-Maclane spectra such an example can't exist since those determined by their homotopy groups...



"Phantom" non-equivalences of spectra?
For this we can use a swindle-type technique.


Let $B = \bigoplus\_{n=2}^\infty H\Bbb Z/2$, and $A = \bigoplus\_{n=2}^\infty \Sigma^{-n} H\Bbb Z/2$. We can construct maps $A \to B$ by specifying their effect on each summand of $A$.


* The first map, $f: A \to B$, is the sum of the composites $$\Sigma^{-n} H\Bbb Z/2 \xrightarrow{Sq^n} H\Bbb Z/2 \to \bigoplus\_{n=2}^\infty H\Bbb Z/2,$$ where the second map includes as the $n$'th summand.
* The second map, $g: A \to B$, is the same, except the we map the $n$'th summand of $A$ to the $(n+1)$'st summand of $B$. (In particular, the summand with $n=2$ is completely missed by $g$.)


Define $X = cofib(f)$ and $Y = cofib(g)$. Since these sums start at $n=2$, $B$ is the connective cover of both $X$ and $Y$ by the long exact sequence in homotopy. I claim that $X$ and $Y$ have equivalent connective covers $\tau\_{\geq m} X$ and $\tau\_{\geq m} Y$ for all $m < 0$, and that they are inequivalent spectra. This suffices to give an example for your question, since taking the associated infinite loop spaces factors through some connective cover.




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First let's show that they have equivalent covers. Note that $\tau\_{\geq m+1} X$ is the cofiber of a map
$$
\bigoplus\_{2 \leq n \leq -m} \Sigma^{-n} H\Bbb Z/2 \xrightarrow{\tau\_{\geq m} f} \bigoplus\_{n=2}^\infty H\Bbb Z/2
$$
and similarly for $\tau\_{\geq m+1} Y$. However, the maps $\tau\_{\geq m} f$ and $\tau\_{\geq m} g$ are equivalent maps in the homotopy category: both are the the same finite list of Steenrod squares summed with countably many extra summands of $H\Bbb Z/2$ on the target. Thus, these connective covers are equivalent.




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Next, let's show that $X$ and $Y$ are not equivalent. To see this, first note that $Y$ has $H\Bbb Z/2$ as a summand: the map $g: A \to B$ completely misses the summand $H\Bbb Z/2$ and allows us to split it off. However, the cofiber sequence defining $X$ gives an exact sequence
$$
[X,H\Bbb Z/2] \to [B, H\Bbb Z/2] \to [A, H\Bbb Z/2]
$$
which is more explicitly
$$
[X,H\Bbb Z/2] \to \prod\_{n=2}^\infty H^0(H\Bbb Z/2) \xrightarrow{\prod Sq^n} \prod\_{n=2}^\infty H^n(H\Bbb Z/2).
$$
This second map is injective, and so by exactness the restriction
$$
[X,H\Bbb Z/2] \to [B, H\Bbb Z/2] \cong Hom(\pi\_0 X, \Bbb Z/2)
$$
must be trivial. Therefore, $X$ can't have $H\Bbb Z/2$ as a summand and thus can't be equivalent to $Y$.


