Note: I will concatenate paths from left to right.
If $X$ is a path-connected topological space, and $p, q\in X$ are distinct, then any path $\gamma$ between $p$ and $q$ induces an isomorphism between $\pi_1(X, p)$ and $\pi_1(X, q)$, where a loop $\alpha$ based at $p$ is send to the loop $\gamma^{-1}\alpha\gamma$, which is based at $q$. If $\delta$ is another path between $p$ and $q$, then $\gamma$ and $\delta$ would induce the same isomorphism iff $\gamma^{-1}\alpha\gamma$ is homotopic (as a loops with fixed base point $q$) to $\delta^{-1}\alpha\delta$ for all $\alpha$, since these would then be in the same equivalence class and thus equal in $\pi_1(X, q)$. This would only seem to hold if every loop from $p$ to $q$ and then back to $p$ in the reverse direction (i.e., all paths of the form $\gamma^{-1}\gamma$ are homotopic.
Now, this condition is enough to answer the question, but I was wondering if it was equivalent to some simpler condition. I guess it's equivalent to all paths from $p$ to $q$ being homotopic to each other (while fixing $p$ and $q$). Is there a name for this condition?
Best Answer
It is equivalent to a simpler condition, but not quite the one you're thinking. Suppose you have two paths $\gamma_1, \gamma_2$ from $p$ to $q$. They induce isomorphisms $\phi_1, \phi_2$. Then the automorphism of $\pi_1(X,p)$ $\phi_2 \circ \phi_1^{-1}$ is just the inner automorphism given by conjugation by the loop $\gamma_1 \gamma_2^{-1}$. Note that any element of $\pi_1(X,p)$ can be realized as such a pair of loops (I leave this tothe reader). So for all isomorphisms to be the same, all inner automorphism of $\pi_1(X,p)$ must be trivial, which is precisely when $\pi_1(X,p)$ is abelian.