= Solution
The nonconstant <invariant factors of a linear operator> satisfy $d_1\mid\cdots\mid d_k=m_A$. By part (a)(i), $m_A=X^2+1$ is an <irreducible polynomial>, so every $d_j$ must equal $X^2+1$. Their <polynomial degrees> sum to $n$, giving $n=2k$.
The <companion matrix> of $X^2+1$, in the <basis> $(1,X)$ of $\mathbb Q[X]/(X^2+1)$, is
$$
C=\begin{pmatrix}0&-1\\1&0\end{pmatrix}.
$$
Therefore, for each fixed even $n$, every such $A$ has <rational canonical form>
$$
\boxed{\operatorname{diag}(\underbrace{C,\ldots,C}_{n/2\text{ blocks}})}.
$$
There are no such <matrices> when $n$ is odd. This is an instance of the <rational canonical form with irreducible minimal polynomial> principle; equality of <minimal polynomials> alone does not determine <rational canonical form> in general.
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