Let $A,B$ be commutative rings in characteristic $p$. Let $\phi_{A}:A\rightarrow A,\phi_{B}:B\rightarrow B$ be the Frobenius morphisms, i.e. the $p$-power maps. If we have ${\rm{colim}}_{n\in\mathbb{N}}A\cong {\rm{colim}}_{n\in\mathbb{N}}B$, where the transition maps are Frobenius morphisms, can we show that $A\cong B$ ?Answer: No. Recall that an $\mathbb{F}_p$-algebra $R$ is perfect if the Frobenius map $\varphi : R \ni r \mapsto r^p \in R$ is an isomorphism. The colimit of powers of the Frobenius you describe is the perfection of an $\mathbb{F}_p$-algebra, and it is named that because it's the left adjoint of the inclusion of perfect $\mathbb{F}_p$-algebras into $\mathbb{F}_p$-algebras. This makes perfect $\mathbb{F}_p$-algebras a reflective subcategory of $\mathbb{F}_p$-algebras, which among other things implies that perfection fixes any perfect $\mathbb{F}_p$-algebra.This is all abstract context for the following more specific counterexample: take $A = \mathbb{F}_p[x]$, whose perfection is $\mathbb{F}_p[x^{\frac{1}{p^{\infty}}}]$, the ring obtained by adjoining all $p^n$-power roots of $x$. Then take $B = \mathbb{F}_p[x^{\frac{1}{p^{\infty}}}]$ to be the perfection. More generally, we could take $A$ to be any $\mathbb{F}_p$-algebra which is not perfect and $B$ to be its perfection.These notes by Bhatt have a more general claim right before Remark 1.4 about when two algebras have isomorphic perfections but I'm not familiar enough with universal homeomorphisms to say anything about it.