I consider two groups of permutation: the group $H$ that acts faithfully on the set $\Delta$ and the group $K$ that acts faithfully on the set $\Omega$. I suppose that $H$ is not the trivial group.
Let be $B=\{f:\Omega \rightarrow H \:|\: supp(f) <\infty \}$ with the operation $(f \cdot f_1)(x)=f(x)f_1(x)$. I have that $(B,\cdot)$ is a group.
I consider this action of $K$ on $B$ defined is this way:
$\forall g \in K $, $\forall f \in B$ and $\forall x \in \Omega$
$$ f^g(x)=f(x^{g^{-1}}).$$
I have proved that this is an action and that : $$ (f f_1)^g=f^g f_1^g.$$
So now I can define the following homomorphism: $$\phi : K \rightarrow Aut(B)$$ where $\phi(g) (f)=f^g$.
My problem: I have difficulties in proving that this homomorphism is injective. How can I deduce that $g=g_1$ from $f(x^{g^{-1}})=f(x^{g_1^{-1}})$ $\forall f \in B$? My idea was to use the fact that the action of $K$ on $\Omega$ is faithful but I don't know how .
Thanks!
Suppose $g\in \ker (\phi)$, so that $f^g=f$ for all $f\in B$. Choose some $h\in H$ such that $h\neq 1_H$. For every $x\in \Omega$ you can define $$f_x:\Omega\to H :y\mapsto \left\{ \begin{array}{ll} h & \text{if }y=x\\ 1_H & \text{otherwise.}\end{array}\right.$$
Then by assumption $f_x^g=f_x$, so in particular. $$f_x(x^{g^{-1}})=f_x(x)=h,$$ for all $x$. This implies that $x^{g^{-1}}=x$ for all $x$, and thus that $g=1_K$.
Thus $\ker(\phi)=\{1_K\}$, which means $\phi$ is injective.