In Spanier's algebraic topology book, in section 1.7 about the fundamental groupoid, he claims that if $A$ and $B$ are objects in the same component of a groupoid $\mathcal{G}$ (meaning that $hom_{\mathcal{G}}(A,B)\neq\varnothing)$, then the set $$\{F(f)\in hom_{Grp}(hom_{\mathcal{G}}(A,A),hom_{\mathcal{G}}(B,B))|f\in hom_{\mathcal{G}}(A, B)\}$$ is a "conjugacy class of isomorphisms in $hom_{Grp}(hom_{\mathcal{G}}(A,A),hom_{\mathcal{G}}(B,B))$", where $F$ is the functor from the groupoid $\mathcal{G}$ to $Grp$, defined in the natural way.
My question is, what does a "conjugacy class of isomorphisms" mean? I know what's a conjugacy class in a group is, but the isomorphisms in $hom_{Grp}(hom_{\mathcal{G}}(A,A),hom_{\mathcal{G}}(B,B))$ don't form a group, at least not canonically.
Perhaps if someone could prove the claim I would understand what is meant by it.
The statement would be better expanded, maybe as follows. Note that Spanier uses the word "equivalence" in a category for what we now call an "isomorphism". Also I will write the composite of $f: A \to B$ and $g:B \to C$ as $fg:A \to C$.
Let $A,B$ be objects of a connected groupoid $G$. Write $G(A)$ for $G(A,A)$; the composition in the groupoid $G$ makes $G(A)$ into a group. Suppose $f \in G(A,B)$. The inverse of $f$ is $f^{-1}: B \to A$ and we can define a "conjugacy" $\chi_f: G(A) \to G(B) $ by $g \mapsto f^{-1}gf$.
If $G$ were a group, then we would have an inner automorphism map $\chi: G \to Aut(G)$. It is possible to generalise this to the case of a groupoid $G$ as follows.
Let $Aut(G)$ be the groupoid with the same objects as $G$ and whose arrows $A \to B$ are all isomorphisms of groups $G(A) \to G(B)$. Then the above conjugacy construction gives a morphism of groupoids $\chi: G \to Aut(G)$.
For more on various uses of groupoids in topology, see this mathoverflow discussion.