 Solution

The matrices are listed in the order $I,a,a^2,a^3$. In the first representation, $D(a)$ is the clockwise quarter-turn matrix and $D(a{)}^2=-I_2$. In the second, $E(a)=i{I}_2$, whose successive powers give the listed scalar matrices. Both therefore obey the group table and distinguish all four elements.

Taking traces gives the characters of a representation:In particular ${\chi }_D(a)=0$ but ${\chi }_E(a)=2i$. Since trace is invariant under similarity, the representations are inequivalent, even after allowing complex changes of basis for the real matrices of $D$.

One can also see the obstruction directly: conjugating the scalar matrix $i{I}_2$ leaves it unchanged, whereas $D(a)$ is not scalar. Over the complex numbers, $D(a)$ has eigenvalues $i,-i$, while $E(a)$ has $i,i$; those different multiplicities cannot be removed by changing basis.