On page 26 of Peter May's A Concise Course on Algebraic Topology, it is claimed that given any two covers of a space $X$, $(E, p)$ and $(E', p')$ are isomorphic iff for any points $e \in E, e' \in E'$ with $p(e) = p'(e')$ then $p(\pi(E, e))$ is conjugate to $p'(\pi(E', e'))$ where $\pi(A, a)$ is the fundamental group of $A$ based at $a$, however the actual thing that seems to follow is that they are isomorphic iff the sub-groups are equal as sets.
I can't figure out how they arrive at the statement about conjugates. Could someone help show me how it follows?
You should look at the Proposition some pages before that, in 3.3 "Coverings of groupoids", which states that as you vary $e \in p^{-1}(b)$, the image of $\pi(\mathcal{E}, e)$ runs through all the different conjugates in $\pi (\mathcal{B}, b)$.
It follows that if you find any two basepoints $e \in E, e \prime \in E \prime$ such that the images are conjugate, like in the theorem, then you can choose possibly a different basepoint of $E$ (or $E \prime$) to make the images equal.
Also, let me just state an opinion that May's presentation of covering theory (in terms of coverings of groupoids) is flawless, as it makes the arguments free from possibly confusing point-set topological clutter.