Continuity of a group homomorphism $\beta\colon G\to\text{Aut}(C_{0}(X))$. Let $G$ be a topological group and $X$ a topological space. Assume that both $G$ and $X$ are locally compact and Hausdorff. Let $G$ act continuously on $X$, that is, we have a (jointly) continuous map $$G\times X\to X,\qquad(t,x)\mapsto t\cdot x$$ such that $e\cdot x=x$ and $(t_{1}t_{2})\cdot x=t_{1}\cdot(t_{2}\cdot x)$, where $e$ denotes the group identity of $G$. Let $\text{Aut}(C_{0}(X))$ be group of all $*$-automorphisms $C_{0}(X)\to C_{0}(X)$. Endow $\text{Aut}(C_{0}(X))$ with the topology of pointwise convergence, that is, $\phi_{i}\to\phi$ if and only if $\phi_{i}(f)\to\phi(f)$ for all $f\in C_{0}(X)$. Together with this topology, $\text{Aut}(C_{0}(X))$ becomes a topological group.
Now consider the map $$\beta\colon G\to\text{Aut}(C_{0}(X)),\qquad\beta(t)(f)(x):=f(t^{-1}\cdot x)$$ for all $t\in G$, $f\in C_{0}(X)$ and $x\in X$.
My question: Why is $\beta$ continuous?
I managed to prove that $\beta$ is a well-defined group homomorphism. Thus it suffices to prove continuity at $e$. More precisely, it suffices to show that $s_{i}\to e$ implies $\|\beta(s_{i})(f)\to f\|_{\infty}\to0$ for all $f\in C_{0}(X)$. Any suggestions on how to continue from here would be appreciated.
 A: The following is not a complete answer, but rather a potential roadmap to obtain the answer. It became too long to post as a comment.

We may assume that $f \in C_c(X)$, i.e. $f$ is compactly supported, because these functions are dense in $C_0(X)$. Indeed, if $f \in C_0(X)$ and $g \in C_c(X)$, we have the estimate
$$\|\beta(s_i)f-f\| = \|\beta(s_i)f-\beta(s_i)g\| + \|\beta(s_i)g-g\| + \|g-f\| \le 2\|f-g\| + \|\beta(s_i)g-g\|.$$

Next, we can invoke the following lemma:
Lemma: If $f \in C_c(X)$, the following condition is satisfied:
For all $\epsilon > 0$, there exists a neighborhood $V \subseteq G$ of $e$ such that $$\sup_{x \in X}|f(y^{-1}x)-f(x)| < \epsilon$$
whenever $y \in V$.
A similar statement can for example be found in the book "A course in abstract harmonic analysis" by Folland on p37-38, if you are looking for inspiration to prove this lemma.

As you already observed, it suffices to show
$$\lim_i \|\beta(s_i)(f)-f\|_\infty = \lim_i\sup_{x \in X}|f(s_i^{-1}x) -f(x)| =0$$
whenever $s_i \to e$.
Using the above lemma, this is easy. Indeed, let $\epsilon > 0$ and choose $V$ as in the lemma. Choose an index $i_0$ such that $s_i \in V$ whenever $i \ge i_0.$ Then $\sup_{x \in X}|f(s_i^{-1}x)-f(x)|< \epsilon$ for all $i \ge i_0$. Hence, the result follows.
