Is there a non trivial sequence $(T\_{n})$ of linear operators $T\_{n}$ on $M\_{n}(\mathbb{C})$ such that $$T\_{nm}(X\otimes Y)=T\_{n}(X) 
\otimes T\_{m}(Y) $$ where $X$ and $Y$ are in $M\_{n}(\mathbb{C})$ and $M\_{m}(\mathbb{C})$, respectively?


By trivial sequence we mean $T\_{n}=Id$ for all $n$.



The functional equation $T(x\otimes y)=T(x)\otimes T(y)$ on the matrix algebra
$T\_n(X) = n \times n$ matrix with all entries $trace(X)$.  

This doesn't satisfy $T\_n(X\_1 X\_2) = T\_n(X\_1) T\_n(X\_2)$.


Edit: More examples:


1. $T\_n(X)$ has $trace(X)$ in upper left corner, 0 elsewhere.
2. $T\_n(X)$ has $trace(X)$ in lower right corner, 0 elsewhere.
3. $T\_n(X)$ has sum of anti-diagonal $(\Sigma x\_{i,n-i})$ in upper right corner, 0 elsewhere.
4. $T\_n(X)$ has sum of anti-diagonal in lower left corner, 0 elsewhere.
5. $T\_n(X)$ has all entries = sum of all entries of $X$.


Edit: Another type:


6. $T\_n(X)=0$ ($n$ even); $T\_n(X)$ has sum of middle row in every middle row entry, 0 elsewhere ($n$ odd).
7. Same sum as in (6.) except $T\_n(X)=0$ has only one non-zero entry at left end resp. middle resp. right end of middle row.


