Here is a very simple claim: if $ G $ is profinite group isomorphic to its completion $ \hat{G} $. Then the natural map $ \phi: G \to \hat{G} $ is an isomorphism. I'm not sure how to show this. The universal property of ($ \hat{G}, \phi $) seems to only give that $ \phi $ is injective. I know that $ G $ is isomorphic to the inverse limit taken over all $ G/N$ with $ N $ open normal subgroups (of finite index), and $ \hat{G} $ is isomorphic to the inverse limit taken over all $ G/K$ taken over all $ K $ normal subgroups of finite index. But it's not so clear how to use this, since the isomorphism is not explicitly given. Help is appreciated.
This claim is taken from these notes, page 6.
This appears to simply be an error in the notes. Certainly, the claim does not follow formally from the universal property as the text seems to imply. Consider the following alternate statement, which has the same formal categorical structure (just replace the forgetful functor adjunction between groups and profinite groups with the forgetful functor adjunction between groups and sets): if a group $G$ is isomorphic to the free group $F(G)$ on its underlying set, then the canonical map $G\to F(G)$ is an isomorphism. This statement is false, though (a free group on any infinite set is a counterexample).
I do not know whether the claim in question is true or not. If it is true, I would expect the proof to involve a hands-on analysis of how $\phi$ can fail to be an isomorphism rather than any formal abstract nonsense.