= 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, $E(a)=iI_2$, 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>:
$$
\boxed{\chi_D=(2,0,-2,0),\qquad\chi_E=(2,2i,-2,-2i).}
$$
In particular $\chi_D(a)=0$ but $\chi_E(a)=2i$. Since trace is invariant under similarity, \b[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 $iI_2$ 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.
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