= Solution
In one dimension the defining relations force
$$
r\mapsto\varepsilon,\qquad t\mapsto\delta,\qquad
\varepsilon,\delta\in\{1,-1\},
$$
giving four characters.
Every irreducible two-dimensional representation is, after a choice of <basis>,
$$
r\mapsto\begin{pmatrix}\lambda&0\\0&\lambda^{-1}\end{pmatrix},
\qquad
t\mapsto\begin{pmatrix}0&1\\1&0\end{pmatrix},
$$
where $\lambda\in\mathbb C^\times\setminus\{1,-1\}$. It is irreducible because the two distinct $r$-eigenlines are interchanged by $t$. Conversely, part (a) gives this <basis> from $v,tv$. If the only <eigenvalue> of $r$ were $\pm1$, its eigenspace would be $t$-invariant and contain a $t$-eigenvector, contradicting irreducibility. Parameters $\lambda$ and $\lambda^{-1}$ give isomorphic representations, and there are no other identifications.
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