Exercise 4.3.1 Etingof confirmed that the book contains an error and the condition that \(f(xi) = \sqrt{-1}f(x)\) should be replaced by \(f(ix) = \sqrt{-1}f(x)\).
Let \(V\) denote the space of functions defined in the exercise. The action of left-multiplication by \(i\) in \(Q_8\) splits the latter into two orbits, \(A = \{1, i, -1, -i\}\) and \(B = \{k, -j, k, j\}\). If \(f \in V\), then \(f\) is uniquely determined by the value it takes on one element from each orbit, say \(1\) and \(k\). Therefore \(V\) is a vector space with dimension 2 and has a basis given by \(f_1, f_2\) where \(f_1(1) = f_2(k) = 1\) and \(f_1(k) = f_2(1) = 0\).
Suppose \(f \in V, g \in Q_8\). Then for each \(x \in Q_8\), we have \(g \circ f(ix) = f(ixg) = \sqrt{-1} f(xg) = \sqrt{-1} g \circ f(x)\), thus \(g \circ f \in V\) also. It is clear that \(1 \circ f = 1\) and \(gh \circ f = g \circ (h \circ f)\). So \(V\) is a valid representation of \(Q_8\).
Direct computation shows that \begin{align*} i \circ f_1 &= if_1 \\ i \circ f_2 &= -if_2 \\ j \circ f_1 &= -if_2 \\ j \circ f_2 &= -if_1 \\ k \circ f_1 &= -f_2 \\ k \circ f_2 &= f_1 \end{align*} so in matrix form, \[ \rho(i) = \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix} \qquad \rho(j) = \begin{pmatrix} 0 & -i \\ -i & 0 \end{pmatrix} \qquad \rho(k) = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \] By inspection, there is no nontrivial common eigenspace of these matrices, so \(V\) is irreducible. (Note that (4.3.1) involves the same set of three matrices, but assigned differently. This can be viewed as a consequence of the fact that \(Q_8\) has an automorphism that sends \(i, j, k\) to \(j, k, i\) respectively.)