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I was asked by a student to give away the answer to the puzzle. The crux of the matter is that any continuous map $$B:S\rightarrow M$$ from a simply connected set $S$ can be used as a base-point for thefundamental group. One way to do this to use $B$ to get a fiber functor$F_B$ that associates to a covering $$N\rightarrow M$$the set of splittings of the covering $$N_B:=S\times_M N\rightarrow S$$ of $S$.If we choosea point $b'\in S$, any splitting is determined by its value at $b'$, givinga bijection of functors$F_B=F_{b'}=F_b$ where $b=B(b')\in M$. Now, when $$B:M'\rightarrow M$$ is the universalcovering space, I will really leave it as a (tautological) exerciseto exhibit a canonical anti-isomorphism$$Aut(F_B)\simeq Aut(M'/M).$$ The 'point' is that $$F_B(M')$$ has a canonicalbase-point that can be used for this bijection.

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I realized I should fix one possible source of confusion. If you work it out, you find that the bijection $$\pi_1(M,b)\simeq M'_b\simeq Aut(M'/M)$$ described above is actually an anti-isomorphism. That is, the order of composition is reversed. Consequently, in the puzzle at the end, the canonical bijection $$Aut(M'/M)\simeq \pi_1(M,B)$$ is an anti-isomorphism as well. However, another simple but amusing exercise is to note that the various bijections with Galois groups, like $$\pi_1(Spec(F), b)\simeq Gal(\bar{F}/F),$$ are actually isomorphisms.
This question reminds me to add another very basic reason to avoid the Galois group as a definition of $\pi_1$. It's rather tricky to work out the functoriality that way, again because the base-point is de-emphasized. In the $Aut(F_b)$ approach, functoriality is essentially trivial.