The functional equation $T(x\otimes y)=T(x)\otimes T(y)$ on the matrix algebra 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 $n$.
 A: $T_n(X) = n \times n$ 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:


*

*$T_n(X)$ has $trace(X)$ in upper left corner, 0 elsewhere.

*$T_n(X)$ has $trace(X)$ in lower right corner, 0 elsewhere.

*$T_n(X)$ has sum of anti-diagonal $(\Sigma x_{i,n-i})$ in upper right corner, 0 elsewhere.

*$T_n(X)$ has sum of anti-diagonal in lower left corner, 0 elsewhere.

*$T_n(X)$ has all entries = sum of all entries of $X$.


Edit: Another type:


*$T_n(X)=0$ ($n$ even); $T_n(X)$ has sum of middle row in every middle row entry, 0 elsewhere ($n$ odd).  

*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.
A: What is a linear operator---do you require $T_n(X_1 X_2) = T_n(X_1) T_n(X_2)$? If not, $T_n(X) = c^n X$ for a fixed scalar $c$ works. Even if yes, I think that $T_n(X) = X$ if $n$ odd, $0$ if $n$ even should work ($T_n = Id$ for odd $n$, but not all $n$).
A: Pick for each prime number $p$ a complex number $\alpha_p$.
For $n$ an integer, define $\alpha_n = \prod \alpha_p ^{\nu_p(n)}$
where $\nu_p(n)$ is the order of $p$ in the prime decomposition of $n$.
One has $\alpha_{nm} = \alpha_n \alpha_m$
Then you can take $T_n(M) = \alpha_n M$
Zach Teitler's example correspond to $\alpha_2=0$ and $\alpha_p =1$ for all odd prime $p$.
