= Solution
Here $a^2=-I$, $a^3=-a$ and $a^4=I$, so $K$ is a <cyclic group of order four>. Its <multiplication table> is
$$
\boxed{\begin{array}{c|rrrr}
\cdot&I&-I&a&-a\\\hline
I&I&-I&a&-a\\
-I&-I&I&-a&a\\
a&a&-a&-I&I\\
-a&-a&a&I&-I
\end{array}}
$$
Write the specified image of $a$ as $A$. Direct matrix multiplication gives
$$
A^2=-I_2,\qquad A^3=-A,\qquad A^4=I_2.
$$
A <group representation> is a <group homomorphism> to a <general linear group>, so the images of all powers of $a$ are forced:
$$
\boxed{\begin{aligned}
\Phi(I)&=\begin{pmatrix}1&0\\0&1\end{pmatrix},\\
\Phi(-I)&=\begin{pmatrix}-1&0\\0&-1\end{pmatrix},\\
\Phi(-a)&=\begin{pmatrix}0&-i\\-i&0\end{pmatrix}.
\end{aligned}}
$$
All four images are invertible. Defining $\Phi(a^j)=A^j$ is well-defined modulo four because $A^4=I_2$, and $A^{j+\ell}=A^jA^\ell$ verifies every product in the table. These observations prove both existence and uniqueness of this <representation> with the prescribed image.
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