If A commutes with both of these matrices, then A must be a scalar multiple of the identity matrix

$A$ commutes with the first matrix implies that $A$ preserves its eigenspaces. This implies that $A(e_i)=c_ie_i,i=1,2,3,4$.

$A$ commutes with the second matrix $C$ implies that $AC(e_1)=A(e_2)=c_2e_2=C(A(e_1)=C(c_1e_1)=c_1e_2$ implies $c_1=c_2$,...

since $AC(e_2)=CA(e_2), AC(e_3)=CA(e_3)$ deduce that $c_1=c_2=c_3=c_4$.

Hint:

Left multiplication of a square matrix by $D=$ first (diagonal) matrix amounts to multiply its rows by the diagonal elements (for this matrix, the first row is multiplied by $1$, the second row by $2$, &c.). Right multiplication amounts to multiplying its columns by the diagonal elements. If both results are equal, by identification, you can deduce that $A$ is a diagonal matrix.

Commutativity of multiplication by the second matrix will then let you show, by identification, that all elements on the diagonal are equal.