Let $\mathcal{G}$ be a coherent 2-group. Following Baez:HDA5 $\mathcal{G}$ is uniquely determined up to 2-equivalence by the following data: The fundamental group $\pi_1(\mathcal{G})=G$, the second homotopy group $\pi_2(\mathcal{G})=H$, the action of $G$ on $H$ by group autos and a class in $a\in H^3(G,H)$. Now theres a third group associated to $\mathcal{G}$ which consists of the morphisms of $\mathcal{G}$ and where multiplication is given by tensor product (horizontal composition in the 2-category). Of course this is only a monoid a priori but becomes a group after quotienting out the cumbersome to state but rather obvious congruence relation defined below.

Let $\epsilon_X:I\rightarrow X\otimes\bar{X}$, $\eta_X:\bar{X}\otimes X\rightarrow I$ be the unit and counit adjoint equivalences. If $f:X\rightarrow Y$ is an arrow in $\mathcal{G}$, there's the so called mate of $f$ denoted $\hat{f}:\bar{X}\rightarrow\bar{Y}$ and defined as $$\hat{f}=(i_{\bar{X}}\otimes\epsilon_Y)(i_{\bar{X}}\otimes f^{-1}\otimes i_{\bar{Y}})(\eta_X\otimes i_{\bar{Y}})$$ Here $i$ is the identity and I use composition from left to right! It is a result of Laplaza that for any object $X$ there's at most one isomorphism $X\cong I$ built out of $\eta$ and $\epsilon$. If there is one, call $X$ simple. Now define $f:X\rightarrow Y$ and $g:A\rightarrow B$ equivalent iff both $X\otimes\bar{A}$ and $Y\otimes\bar{B}$ are simple and the composition $$I\rightarrow X\otimes\bar{A}\xrightarrow{f\otimes\hat{g}}{}Y\otimes\bar{B}\rightarrow I$$ is the identity.

Now from the above classification of 2-groups it should be possible to construct this group out of $G,H$ and $a$. Does anybody know how this operation on the data $G,H,a$ looks like? Is it a well known construction?