The argument on page 105, starting from the paragraph “We define a homomorphism…” may be harder to follow than we thought, and so let us give more details here.
Given $H\leq \mbox{Sym}(\Delta)$ and $B=\mbox{Func}(\Delta, G)$, define a map $\tau:H\to \mbox{Aut}(B)$ as follows: for $f\in B$, $f(h\tau)$ is the map
\[
f(h\tau): \delta\mapsto (\delta h^{-1})f,\quad\mbox{or equivalently}\quad \delta(f(h\tau))= (\delta h^{-1})f.
\]
To prove that $f(h\tau)$ lies in $\mbox{Aut}(B)$ is routine but it involves proving that $h\tau$ is a bijection, and that $(f_1f_2)(h\tau)=(f_1(h\tau))(f_2(h\tau))$, noting that multiplication in $B$ is pointwise.
To prove that $\tau$ is a homomorphism, take $h, h’\in H$. Then $(h\tau)\circ (h’\tau)$ maps an arbitrary $f\in B$ to $f_1(h’\tau)$, where $f_1=f(h\tau)$. (Note that $\circ$ denotes composition of automorphisms in $\mbox{Aut}(B)$.) Using the display above we have
\begin{align*}
\delta(f((h\tau)\circ (h’\tau)))&= \delta(f_1((h’\tau)))\\
&=(\delta (h’)^{-1})f_1 \\
%=(\delta (h’)^{-1})(f(h\tau)) \\
&= (\delta (h’)^{-1}h^{-1})f \\
&= (\delta (hh’)^{-1})f\\
&=\delta(f((hh’)\tau).
\end{align*}
Since this holds for all $\delta\in\Delta$ we have $f((h\tau)\circ (h’\tau)) = f((hh’)\tau)$, and since this holds for all $f\in B$ we have $(h\tau)\circ (h’\tau) = (hh’)\tau$. Hence $\tau$ is a homomorphism.