I'm quite sure I am correct about this but at the moment I can't think for the life of me why. Suppose $A$ and $B$ are $*$-algebras and there are $*$-homomorphisms $\pi_1 \colon A \to \mathcal{B}(\mathcal{H}_1)$ and $\pi_2 \colon B \to \mathcal{B}(\mathcal{H}_2)$ and a unitary map $U \colon \mathcal{H}_1 \to \mathcal{H}_2$ such that $U \pi_1(A) U^* = \pi_2(B)$.
Does this imply that the $C^*$-algebras $\overline{\pi_1(A)}^{\mathcal{B}(\mathcal{H_1})}$ and $\overline{\pi_2(B)}^{\mathcal{B}(\mathcal{H}_2)}$ are $*$-isomorphic?
I think your unitary goes from $\mathcal H_2\to\mathcal H_1$, or equivalently is an element in the set of intertwiners $(\mathcal B(\mathcal H_2),\mathcal B(\mathcal H_1))$. Given this, the unitary in question allows you to identify operators between those two Hilbert spaces, i.e. it induces a bijection through its adjoint action. Any Cauchy net in $\pi_1(A)$, which converges in $\overline{\pi_1(A)}$ by completeness, is then mapped to a Cauchy net in $\pi_2(B)$, which converges in $\overline{\pi_2(B)}$ for the same reason. A moment's thought shows you that these limits, say $x$ and $y$ respectively, are linked by the unitary $U$, i.e. $y = UxU^*$ and that the boundary is mapped onto the boundary. All this is then enough to conclude that those two C*-algebras are $*$-isomorphic.