Suppose $A_i$ is a matrix algebra for $i=1,...,n$ (i.e., $A_i$ consists of $n_i\times n_i$ matrices with complex entries), and suppose $\tau$ and $\sigma$ are elements of $A_1\otimes \cdots \otimes A_n$ satisfying the following properties:
$\tau$ and $\sigma$ are both Hermitian.
$\text{tr}_{A_1\otimes \cdots \otimes \widehat{A_i}\otimes \cdots A_n}(\tau)$ and $\text{tr}_{A_1\otimes \cdots \otimes \widehat{A_i}\otimes \cdots A_n}(\sigma)$ are both invertible for $i=1,...,n-1$ ($\widehat{A_i}$ denotes the empty algebra and $\text{tr}_{(*)}$ is the partial trace).
$\text{tr}_{A_1\otimes \cdots \otimes\widehat{A_{i-1}}\otimes \widehat{A_{i}}\otimes\cdots \otimes A_{n}}(\sigma)=\text{tr}_{A_1\otimes \cdots \otimes\widehat{A_{i-1}}\otimes \widehat{A_{i}}\otimes\cdots \otimes A_{n}}(\tau)$ for $i=2,...,n$.
Then can we conclude that $\tau=\sigma$?