Can ($X^I$, product topology) and ($X^I$, box topology) be homeomorphic for some nontrivial $X$ and infinite $I$? This is a verbatim repost of this question by Jianing Song. A few months ago I placed a bounty on the question but there were no answers, so I am reposting it here.

Let $X$ be a nontrivial topological space, $I$ be a infinite set, we can endow $X^I$ (the set of all functions $I\to X$) with either the product topology or the box topology. We know that the box topology is strictly finer than product topology, but since a topology can be homeomorphic to a strictly finer topology, can the product topology be homeomorphic to the box topology anyway?
EDIT: Here I interpret the term "nontrivial" by that the topology endowed on $X$ is not the trivial topology. I do not ask that $X$ is Hausdorff, but I think restricting the problem to Hausdorff spaces is also a question worth asking.
 A: This answer was supposed to contain a counterexample, but there was a mistake in it. In any case I think the following class of topological spaces may be useful to find a counterexample:
Suppose that we have a set $X$ and a collection $\mathcal{A}$ of subsets of $X$ closed under finite unions. Then we can define a topology on $2^X$ by the following basis of clopens: let $A\in\mathcal{A}$ and let $g:A\to\{0,1\}$ be a function. Then we define $C_{A,g}=\{f:X\to\{0,1\};f\text{ is an extension of }g\}$. Then $\{C_{A,g};A\in\mathcal{A},g:A\to\{0,1\}\}$ is a basis for a topology on the set $2^X$.
For example, $2^\mathbb{N}$ is obtained by setting $X=\mathbb{N}$ and $\mathcal{A}=\{$ finite subsets of $\mathbb{N}\}$. Similarly, $(2^\mathbb{N})^\mathbb{N}_\text{box}$ can be obtained by setting $X=\mathbb{N}^2$, and $\mathcal{A}=\{A\subseteq\mathbb{N}^2;A\cap(\mathbb{N}\times\{n\})\text{ is finite for every }n\in\mathbb{N}\}$, and $(((2^\mathbb{N}_\text{box})^\mathbb{N})^\mathbb{N}_\text{box})^\mathbb{N}$ can be expressed by setting $X=\mathbb{N}^4$ and saying that a set $A$ is in $\mathcal{A}$ if $\{a_1\in\mathbb{N};(a_1,a_2,a_3,a_4)\in A\}$ is bounded and for every pair $(x_1,x_2)$, the set $\{a_3\in\mathbb{N};(x_1,x_2,a_3,a_4)\in A\}$ is bounded.
In general, if we let $2^X_\mathcal{A}$ be the space we defined, then for any infinite set $I$:

*

*$(2^X_\mathcal{A})^I_\text{box}\cong 2^Y_\mathcal{B}$, where $Y=X\times I$ and $\mathcal{B}=\{B\subseteq X\times I;B\cap (X\times\{i\})\in\mathcal{A}\;\forall i\in I\}$. (Here we identified $X\times\{i\}$ with $X$)


*$(2^X_\mathcal{A})^I\cong 2^Y_\mathcal{B}$, where $Y=X\times I$ and $\mathcal{B}=\{B\subseteq X\times I;B\cap (X\times\{i\})\in\mathcal{A}\;\forall i, B\cap (X\times\{i\})=\varnothing\text{ for all $i$ except finite}\}$.
Moreover, suppose we have two spaces $2^X_\mathcal{A}$ and $2^Y_\mathcal{B}$ and we have a bijection $\phi:X\to Y$ which induces a bijection $\mathcal{A}\to\mathcal{B}$. Then the function $F:2^X_\mathcal{A}\to2^Y_\mathcal{B};F(f)=f\circ\phi^{-1}$ is a homeomorphism, just by how we have defined the topologies.
