Let $A\in End(\mathbb{C}^n)$ be Hermitian and with positive eigenvalues. Define the linear map $T:End(\mathbb{C}^n)\longrightarrow End(\mathbb{C}^n)$ with $T(M)=(tr(M))A$. How can I make an isometry $U\in Hom(\mathbb{C}^n,\mathbb{C}^n\times\mathbb{C}^n)$ such that $$T(M)=tr_B(UMU^H)$$ where $tr_B(N\otimes P)=(tr(N))P$ and $U^H$ is the Hermitian conjugation of $U$.
Just an algorithmic step-by-step receipt for computing this $U$ is my question not it's theoretical possibility.
The partial trace is not a standard feature of the Stinespring decomposition. I don't think you can get the decomposition you want in general.
Write $$ U=\begin{bmatrix}V_1\\ \vdots \\ V_n\end{bmatrix}. $$ Then $$ UMU^H=\begin{bmatrix}V_1MV_1^*&V_1MV_2^*&\cdots&V_1MV_n^*\\ V_2MV_1^*&V_2MV_2^*&\cdots&V_2MV_n^*\\ \vdots \\ V_nMV_1^*&V_nMV_2^*&\cdots &V_nMV_n^* \end{bmatrix}, $$ and $$ \text{tr}_B(UMU^H)=\begin{bmatrix}\text{tr}(V_1MV_1^*)&\text{tr}(V_1MV_2^*)&\cdots&\text{tr}(V_1MV_n^*)\\ \text{tr}(V_2MV_1^*)&\text{tr}(V_2MV_2^*)&\cdots&\text{tr}(V_2MV_n^*)\\ \vdots \\ \text{tr}(V_nMV_1^*)&\text{tr}(V_nMV_2^*)&\cdots &\text{tr}(V_nMV_n^*) \end{bmatrix}. $$ You want this last matrix to be equal to $\text{tr}(M)A$ for all $M$. That is, $$\tag{1} \text{tr}(MV_j^*V_k)=\text{tr}(M)A_{kj} $$ for all $M$. Taking $M=E_{kj}$ (matrix unit) we see that $\text{tr}(V_j^*V_k)=0$ if $k\ne j$. But now, taking $M=I$, we deduce that $A_{kj}=0$ when $k\ne j$, i.e. $A$ has to be diagonal. Now we have $\text{tr}(MV_j^*V_k)=0$, when $j\ne k$, for all $M$; it follows that $V_j^*V_k=0$.
From $U^HU=I_n$ we get $$\tag{2}V_1^*V_1+\cdots+V_n^*V_n=I_n.$$ Now adding over $k=j$ on $(1)$ we obtain $$ \text{tr}(M)=\text{tr}(M)\,\sum_kA_{kk}, $$ so $\text{tr}(A)=1$.
Fix $k$. Going back to $(1)$, we have $$ \text{tr}(MV_k^*V_k)=\text{tr}(M)A_{kk} $$ for all $M$. As the right-hand-side does not change as we choose $M=E_{11},\ldots,E_{nn}$, we get that the diagonal of $V_k^*V_k$ is constant, equal to $A_{kk}$. Taking $M=E_{st}$ for any $s\ne t$ we get $$ (V_k^*V_k)_{st}=\text{tr}(E_{1t}V_k^*V_kE_{s1})=\text{tr}(E_{st}V_k^*V_k)=\text{tr}(E_{st})A_{kk}=0. $$ Thus, $V_k^*V_k=A_{kk}I_n$. As $A_{kk}>0$, we deduce that the matrix $V_k/A_{kk}^{1/2}$ is a unitary. But now we get that, for $j\ne k$, $$ 0=V_j^*V_k $$ is a scalar multiple of a unitary: a contradiction. In conclusion, no such $U$ can exist.