For what conditions will two paths not induce the same isomorphism? A problem in my book states:

Under what conditions will two path classes, $\gamma$ and $\gamma'$, from $x$ to $y$ give rise to the same isomorphism of $\pi(X,x)$ onto $\pi(X,y)$?

For the two isomorphisms to be the same, then we need:
$$\forall\beta \in \pi(X,x) : \gamma^{-1}\beta\gamma = (\gamma')^{-1}\beta(\gamma')$$
$$\forall \beta \in \pi(X,x) : \gamma' \gamma^{-1} \beta = \beta \gamma' \gamma^{-1}$$
Hence $\gamma'\gamma^{-1}$ commutes with every element $\beta$. This doesn't seem very satisfactory though, is there a geometrical interpretation of this condition? What does it mean for this path to commute?
 A: The algebraic model of this situation that one needs for algebraic topology is that of groupoids, which can be found developed in these downloadable books: Categories and Groupoids, Topology and Groupoids, so I'll assume you know the basic definitions, and relation with the funamental groups and  groupoid. 
Here is something which is not in these books!   Let $G$ be a groupoid. There is a groupoid $AUT(G)$ whose objects are the same as those of $G$ and whose arrows $x \to y$ are the isomorphisms of object groups $G(x) \to G(y)$.There is a morphism of groupoids $\chi: G \to AUT(G)$ which sends $a: x \to y$ in $G$ to the conjugation isomorphism $G(x) \to G(y) $ given by $u \mapsto a^{-1}ua$. (There is a question   of the order in which one writes this and the conventions in these two books differ, but I'll leave you to draw the pictures and sort it out. The point is:  does one write the composite (product) in $G$ of $a: x\to y$ and $b: y \to z$ as $ab$ or $ba$?) 
So you can now work out that: if $a,b: x \to y$ in $G$ then $\chi(a)= \chi(b)$ if and only if  $ba^{-1} u= u ba^{-1}$ for all $u \in G(x)$. 
