It says the following.

Let us keep the notations of section $3.1$. In addition to choosing local objects $x_i\in \mathcal{P}(U_i)$, in a gerbe $\mathcal{P}$ on $X$, we now choose arrows $x_j\rightarrow x_i$ in $\mathcal{P}(U_{ij})$. Since $G_i=\underline{\text{Aut}}(x_i)$, a chosen arrow $\phi_{ij}$ induces by conjugation a homomorphism of group bundles $$G_j|_{U_{ij}}\xrightarrow{\lambda_{ij}}G_i|_{U_{ij}}$$
with $\lambda\mapsto \phi_{ij}\gamma\phi_{ij}^{-1}$.

All that I know is, as $\mathcal{P}$ is a stack, $\underline{\text{Aut}}(x_i)$ is a sheaf of groups. There is a bijective correspondence between sheaf of groups on $X$ and group bundles(Etale spaces) on $X$. So, I was thinking I have to see $G_i$ as a group bundle on $U_i$ i.e., as a map $G_i\rightarrow U_i$ and then try to understand the homomorphism of group bundles. This is only because they have said **homomorphism of group bundles**. But, I realized it is easy and equivalent to see $\lambda_{ij}$ as a morphism of sheaves itself. It seems unnecessary and confusing as well.

So, we try to understand $\lambda_{ij}$ as a morphism of sheaves. sheaves are in bijective correspondence with group bundles so morphism of sheaves are in bijective correspondence with morphism of sheaves. So, it is just the same.

Now the question is,

Does it make sense to write
$\gamma\mapsto \phi_{ij}\gamma\phi_{ij}^{-1}$ for morphism of sheaves $\lambda_{ij}:G_i|_{U_{ij}}\rightarrow G_j|_{U_{ij}}$?

Answer is, **by abuse of notation**, Yes.

Let $\mathcal{F},\mathcal{G}$ be sheaves on $X$. By a morphism of sheaf $\eta:\mathcal{F}\rightarrow \mathcal{G}$, we mean
collection of maps $\eta(U):\mathcal{F}(U)\rightarrow \mathcal{G}(U)$ that are compatible with restriction maps on respective sheaves.

So, considering shevaes $G_i|_{U_{ij}}, G_j|_{U_{ij}}$ on $U_{ij}$, we define a morphsim $\lambda_{ij}:G_j|_{U_{ij}}\rightarrow G_i|_{U_{ij}}$ of sheaves
. For that, we need to give a map
$\lambda_{ij}(V):G_j|_{U_{ij}}(V)\rightarrow G_i|_{U_{ij}}(V)$ for each $V\subseteq U_{ij}$,
this is just the same as $\lambda_{ij}(V):G_j(V)\rightarrow G_i(V)$.

To make sense of
$G_i(V)$ and $G_j(V)$ we fix up some notation.

We have inclusions $\rho:V\rightarrow U_{ij},\alpha_i:U_{ij}\rightarrow U_i,
\alpha_j:U_{ij}\rightarrow U_j$.

Then, by definition, $G_j(V)=\text{Hom}((\alpha_j\circ \rho)^*(x_j),(\alpha_j\circ \rho)^*(x_j))$.

Let $\eta:(\alpha_j\circ \rho)^*(x_j)\rightarrow (\alpha_j\circ \rho)^*(x_j)$ be an arrow in $\mathcal{P}(V)$. This is same thing as
$$\eta:\rho^*(\alpha_j^*(x_j))\rightarrow \rho^*(\alpha_j^*(x_j))$$ in $\mathcal{P}(V)$. As $\alpha_j^*(x_j)$ is just $x_j|_{U_{ij}}$, we have
$$\eta:\rho^*(x_j|_{U_{ij}})\rightarrow \rho^*(x_j|_{U_{ij}})$$ in $\mathcal{P}(V)$.

The map $\phi_{ij}:x_j|_{U_{ij}}\rightarrow x_i|_{U_{ij}}$ in $\mathcal{P}(U_{ij})$ gives an arrow $\rho^*(\phi_{ij}):\rho^*(x_j|_{U_{ij}})\rightarrow \rho^*(x_i|_{U_{ij}})$ in $\mathcal{P}(V)$. The composition
$$\rho^*(x_i|_{U_{ij}})\xrightarrow{\rho^*(\phi_{ij})}\rho^*(x_j|_{U_{ij}})\xrightarrow{\eta} \rho^*(x_j|_{U_{ij}})\xrightarrow{\rho^*(\phi_{ij})^{-1}}\rho^*(x_i|_{U_{ij}})$$
gives an arrow $\rho^*(x_i|_{U_{ij}})\rightarrow \rho^*(x_i|_{U_{ij}})$.

As $x_i|_{U_{ij}}=\alpha_i^*(x_i)$, this gives an arrow $\rho^*(\alpha_i^*(x_i))\rightarrow \rho^*(\alpha_i^*(x_i))$, which is same as
$(\alpha\circ \rho)^*(x_i)\rightarrow (\alpha\circ \rho)^*(x_i)$ an element in $G_i(V)=\text{Hom}((\alpha_i\circ \rho)^*(x_i),(\alpha_i\circ \rho)^*(x_i))$.

This gives map $G_j(V)\rightarrow G_i(V)$ given by $\eta\mapsto \rho^*(\phi_{ij})^{-1}\circ \eta\circ \rho^*(\phi_{ij})$. This by abuse of notation, can be denoted by $\eta\mapsto \phi_{ij}^{-1}\eta\phi_{ij}$.