Obviously, $\min trace(AX)=-1/2$ for $X=(1/2) I$. Note $A + A^T=A +A^* \ge 0$, so $AX+X A^T \le 0$. If $A$ diagonal, then this $X$ is obviously optimizes the minimum, as $X\le (1/2) I$ is your constraint.

If $A'$ is not is diagonal, write $A'=U A U^*$, $trace(UAU^* UXU^*)=trace(AX)$.
And $UXU^*\le U(1/2)IU^*=(1/2)I$. And $UAXU^* + UXA^*U^* \le 0 \Leftrightarrow AX +X A^* \le 0$, so constraints are invariant under transformation $U$. Take optimum $U X U^*$.