Extending $*$-isomorphisms between $*$-algebras to cross products.

Question

Let $G$ be a discrete countable group and suppose I have two $G$-$C^*$-algebras $A$ and $B$ such that there exists a $G$-equivariant isometric $*$-isomorphism $\varphi \colon A \to B$. One can extend this to a $G$-equivaraint $*$-isomorphism between $\overline{\varphi} \colon C_c(G,A) \to C_c(G,B)$ where $$ \overline{\varphi} (\sum_{g \in G} f_g \delta_g) = \sum_{g \in G} \varphi(f_g)\delta_g. $$

Suppose there exists two covariant representations $(u_1, \sigma_1, \mathcal{H})$ and $(u_2, \sigma_2, \mathcal{H})$ of $C_c(G,A)$ and $C_c(G,B)$ respectively into the same Hilbert space $\mathcal{H}$. Does there exist a unitary $U \colon \mathcal{H} \to \mathcal{H}$ such that $$ U \sigma_1(f)U^* = \sigma_2( \overline{\varphi}(f)) $$ for all $f \in C_c(G,A)$? Further would this mean that the cross products are isomorphic as $C^*$-algebras?

Answer 1

Consider the case when $G=\{e\}$ is trivial, $A=B$ and $\varphi=Id_A.$ Then $C_c(G,A)\simeq A.$ The positive answer to your question would mean that every two representations of $A$ on the same Hilbert space are unitarily equivalent, which is of course false.