Solution: We will find the matrix of $T$ respect to the standard basis of $M_2(\mathbb{C} )$. Recall that the standard basis of $M_2(\mathbb{C} )$ is given by \[ \mathcal{B} = \left\{ e_{11}, e_{12} , e_{21} , e_{22} \right\}, \] where $e_{ij} \in M_2(\mathbb{C} )$ whose $ij^{\text{th} }$ entry is $1$ and other entries are $0$. We have \begin{align*} T(e_{11} ) & = Ae_{11} = \begin{pmatrix} 0 & 2 \\ 2 & 0 \\ \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 0 \\ \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 2 & 0 \\ \end{pmatrix} = 2e_{21} \\[1ex] T(e_{12} ) & = Ae_{12} = \begin{pmatrix} 0 & 2 \\ 2 & 0 \\ \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 0 & 0 \\ \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 2 \\ \end{pmatrix} = 2e_{22} \\[1ex] T(e_{21} ) & = Ae_{12} = \begin{pmatrix} 0 & 2 \\ 2 & 0 \\ \end{pmatrix} \begin{pmatrix} 0 & 0 \\ 1 & 0 \\ \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 0 & 0 \\ \end{pmatrix} = 2e_{11} \\[1ex] T(e_{22} ) & = Ae_{12} = \begin{pmatrix} 0 & 2 \\ 2 & 0 \\ \end{pmatrix} \begin{pmatrix} 0 & 0 \\ 0 & 1 \\ \end{pmatrix} = \begin{pmatrix} 0 & 2 \\ 0 & 0 \\ \end{pmatrix} = 2e_{12} \\[1ex] \end{align*} So, the matrix of $T$ with respect to the standard basis is \[ T = [T]_\mathcal{B} = \begin{pmatrix} 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 2 \\ 2 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ \end{pmatrix}. \] Now the characteristic polynomial of $T$ can be find by $\det (T - \lambda I) = 0$. \begin{align*} \det (T - \lambda I) = 0 & \implies \begin{vmatrix} -\lambda & 0 & 2 & 0 \\ 0 & -\lambda & 0 & 2 \\ 2 & 0 & -\lambda & 0 \\ 0 & 2 & 0 & -\lambda \\ \end{vmatrix} = 0 \\[1ex] & \implies -\lambda (-\lambda (\lambda ^2) + 2(2\lambda )) + 2(-2(\lambda ^2 - 4)) = 0 \\ & \implies \lambda ^4 - 8\lambda ^2 + 16 = 0. \end{align*} Hence, the characteristic polynomial of $T$ is \[ X^4 - 8X^2 + 16 = 0. \]

(An idea from math.stackexchange leads to an easier method).

This can also be solved by observing that $A$ is diagonalizable with eigenvalues $2$ and $-2$. Let \[ A u_j = \lambda _j u_j \quad \text{ and } \quad v_j^{\ast} A = \lambda _j v_j^{\ast} \] for some basis $u_j$ and for some dual basis $v_j^{\ast} $. Then note that $\{ u_j v_k^{\ast} \} $ forms a basis for $M_2(\mathbb{C} )$. Also, \[ T\left( u_j v_k^T \right) = A \left( u_j v_k^T \right) = \lambda _j \left( u_j v_k^T \right). \] Thus, $u_j v_k^T$ is an eigenvector with eigenvalue $\lambda _j$. Hence the characteristic polynomial of $T$ is given by \begin{align*} (x - 2)(x-2)(x+2)(x+2) & = (x-2)^2 (x+2)^2 \\ & = (x^2 - 4)^2 \\ & = x^4 - 8x^2 + 16 = 0. \end{align*}