‘Non-Induced’ Left Regular Representations of $C^{*}$-Dynamical Systems

In what follows, a ‘$*$-representation’ always means a non-degenerate $*$-representation.

Let $(\mathscr{A},G,\alpha)$ be a $C^{*}$-dynamical system, and let $\pi: \mathscr{A} \to B(\mathcal{H})$ be a faithful $*$-representation of $\mathscr{A}$ on a Hilbert space $\mathcal{H}$. Define an associated faithful $*$-representation $\tilde{\pi}: \mathscr{A} \to B({L^{2}}(G,\mathcal{H}))$ and a unitary group representation $\lambda: G \to U({L^{2}}(G,\mathcal{H}))$ as follows.

For all $a \in \mathscr{A}$, $g \in G$, $x \in G$ and $\xi \in {L^{2}}(G,\mathcal{H})$, \begin{align*} [[\tilde{\pi}(a)](\xi)](x) & \stackrel{\text{def}}{=} [\pi({\alpha_{x^{-1}}}(a))](\xi(x)), \\ [[\lambda(g)](\xi)](x) & \stackrel{\text{def}}{=} \xi(g^{-1} x). \end{align*}

$\lambda$ is usually called the left regular representation of $G$.

It is well-known that the pair $(\tilde{\pi},\lambda)$ — called the left regular representation induced by $\pi$ — is a covariant representation of $(\mathscr{A},G,\alpha)$ on ${L^{2}}(G,\mathcal{H})$ and that the integrated form $\tilde{\pi} \rtimes \lambda$ is a faithful $*$-representation of the twisted convolution $*$-algebra $({C_{c}}(G,\mathscr{A}),\star,^{*})$ on ${L^{2}}(G,\mathcal{H})$.

It is also well-known that $\tilde{\pi} \rtimes \lambda$ extends uniquely to a faithful $*$-representation of the reduced crossed product $C^{*}$-algebra $\mathscr{A} \rtimes_{\alpha,\text{r}} G$ on ${L^{2}}(G,\mathcal{H})$.

My question: Suppose that we have a faithful $*$-representation $\rho: \mathscr{A} \to B({L^{2}}(G,\mathcal{H}))$ that is not associated to $\pi$, i.e. is not of the form $\tilde{\pi}$, for any faithful $*$-representation $\pi: \mathscr{A} \to B(\mathcal{H})$. Suppose further that the pair $(\rho,\lambda)$ — which I shall call a non-induced left regular representation — is a covariant representation of $(\mathscr{A},G,\alpha)$ on ${L^{2}}(G,\mathcal{H})$. Then is it necessarily true that the integrated form $\rho \rtimes \lambda$ gives rise to a faithful $*$-representation of $\mathscr{A} \rtimes_{\alpha,\text{r}} G$ on ${L^{2}}(G,\mathcal{H})$?

Thank you very much for your help!

The answer is no in two ways. Let $G$ be a discrete group and consider the trivial action of $G$ on the reduced group $\mathrm{C}^*$-algebra $\mathrm{C}^*_{\mathrm{r}}G$. The pair of the right regular representation $\rho\colon \mathrm{C}^*_{\mathrm{r}}G\to B(\ell_2G)$ and the left regular representation $\lambda\colon G\to U(\ell_2G)$ is covariant. The resulting $*$-homomorphism $\rho\rtimes\lambda$ is continuous on the reduced crossed product (which is nothing but the minimal tensor product $\mathrm{C}^*_{\rho}G \otimes_{\min} \mathrm{C}^*_{\lambda}G$) if and only if $G$ is amenable. It is faithful on the algebraic tensor product $\mathrm{C}^*_{\rho}G \otimes_{\mathrm{alg}} \mathrm{C}^*_{\lambda}G$ if and only if $G$ is ICC.