Solution: Note that the hypothesis that $\mathop{\mathrm{tr}}(AX)=0 $ is true for any traceless matrix $X$. So our strategy to solve this problem will be to choose different traceless matrix $X$ so that we can find the elements of the matrix $A$ using the given hypothesis.

Let $E_{ij}$ denotes the matrix of size $n$ with $ij^{\text{th}}$ entry being zero. For $i\neq j$, take $X=E_{ij}$. Then we have \begin{align*} \mathop{\mathrm{tr}}(AX) = 0 & \implies \mathop{\mathrm{tr}}(AE_{ij}) =0 \\ & \implies a_{ji} = 0. \end{align*} So we prove that $A$ is a diagonal matrix with entries $a_{ii}, ~ i=1,2,\ldots n$, i.e., \[ A = \mathop{\mathrm{diag}}(a_{11},a_{22},\cdots,a_{nn}). \]

Now it remains to show that $a_{11}=a_{22}=\cdots=a_{nn}$. Consider the matrix \[ E_i = \mathop{\mathrm{diag }}(1,0,\cdots,-1,0,\cdots,0), \] where $-1$ is placed at the $ii^{\text{th}}$ place. If we take $X=E_i$, then \begin{align*} \mathop{\mathrm{tr}}(AX) = 0 & \implies \mathop{\mathrm{tr}}(AE_{i}) =0 \\ & \implies a_{11} - a_{ii} = 0 \\ & \implies a_{11} = a_{ii}. \end{align*} Thus, $A = \mathop{\mathrm{diag }}(a_{11},a_{11},\cdots,a_{11})$ and hence, we proved that $A=a_{11}I$.