Consider two Hilbert spaces $\mathcal{H}_1=\mathbb{C}^{k}$, $1\leq k$ and $\mathcal{H}_2=\mathcal{H}_1\otimes\mathbb{C}^{n}$ $1\leq n$. Let $\mathbb{L}(\mathcal{H}_{1,2})$ denote the space of linear operators acting on them.
Now, let $R:\mathbb{L}(\mathcal{H}_{1}) \rightarrow \mathbb{L}(\mathcal{H}_{2})$ be a $*$-homomorphism. One example is the trivial extension by identity operator, i.e., $$ R_0: X \mapsto X\otimes \mathbb{1}_{n}\,,\quad X\in \mathbb{L}(\mathcal{H}_{1}) $$ My question is whether all $*$-homomorphisms are unitarily equivalent to the trivial one? More concretely, given an arbitrary $R:X\mapsto R[X]$, whether there exists a unitary transformation $U\in \mathbb{L}(\mathcal{H}_2)$ such that $UR[X]U^\dagger = X\otimes \mathbb{1}_n$ for all $X\in\mathbb{L}(\mathcal{H_1})$.
Any comments and suggests on references are appreciated!
In finite dimension, yes, if you assume that your homomorphism is unital.
First note that since $L(\mathcal H_1)$ is simple, every $*$-homomorphism is injective and hence isometric. If you now think of the matrix units $E_{kj}$ in $L(\mathcal H_1)=M_k(\mathbb C)$, the $*$-monomorphism maps them to an incomplete $k\times k$ set of matrix units in the codomain. Then $RE_{11},\ldots,RE_{kk}$ are equivalent projections that add to the identity. Then $n=\operatorname{Tr}(E_{11})$ and you can think of $RE_{kj}$ as $E_{kj}\otimes 1_n$ (the "think of" will give you the unitary).