The problem is suppose $M_2(R)$ is matrix ring over real numbers and $T,S\colon M_2(R) \to M_2(R)$ linear bijective transformations (i.e. maps). And suppose the following are hold: $T(x)\cdot S(x) = S(x)\cdot T(x)$ and $T^2(x)\cdot S^2(x) = S^2(x)\cdot T^2(x)$ for all $x\in M_2(R)$, where "$\cdot$" the usual matrix multiplication and $T^2(x)$ means composition of $T$, i.e. $T^2(x)=T(T(x))$. Prove that there exist $\lambda\in R$ such that $T(x)=\lambda S(x)$ for all $x\in M_2(R)$.
My attempt. Put $P = ST^{-1}$-bijective linear transformation. Then we have $P(x)\cdot x = x\cdot P(x)$ for all $x\in M_2(R)$. Suppose $e_{1}= \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$, $e_{2}= \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$, $e_{3}= \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$, $e_{4}= \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$ standart basis of $M_2(R)$ and $P(e_k)=\alpha_{k1}e_1+\alpha_{k2}e_2+\alpha_{k3}e_3+\alpha_{k4}e_4$, $k=1,2,3,4$ and $\alpha_{ij}\in R$. Then using the fact $P(e_k)\cdot e_k = e_k\cdot P(e_k)$ I obtain that $\alpha_{12}=\alpha_{13}=\alpha_{23}=\alpha_{32}=\alpha_{42}=\alpha_{43}=0$, $\alpha_{21}=\alpha_{24}$, $\alpha_{31}=\alpha_{34}$ $\alpha_{22}=\alpha_{33}$ and $\alpha_{11}-\alpha_{14}=\alpha_{44}-\alpha_{41}$. But that's not enough to show that $P(x)=\lambda x$ for all $x\in M_2(R)$. In fact if we consider the following linear transformation $P(\begin{pmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{pmatrix}) = \begin{pmatrix} 2x_{11}+x_{12}-x_{21}+x_{22} & x_{12} \\ x_{21} & x_{11}+x_{12}-x_{21}+2x_{22} \end{pmatrix} $ then we have $P(x)\cdot x = x\cdot P(x)$ for all $x\in M_2(R)$. And here I realized that I have not use the fact that $T^2(x)\cdot S^2(x) = S^2(x)\cdot T^2(x)$ for all $x\in M_2(R)$. But if we put $Q = S^2T^{-2}$ than we don't know that $Q=P^2$ and the argument above doesn't work. So how to solve the problem? Any help is appreciated.
The statement is not true. Counterexample: $T(x)=x$ and $S(x)=x+\operatorname{tr}(x)I_2$.