 Fricke transform of a primitive coprime twist

Use the determinant-normalized slash operator and $W_M=\left(\begin{array}{cc}0& -1\\ M& 0\end{array}\right)$. For units $u,v$ with $Nuv\equiv -1(\mathrm{mod}D)$, the matrix identity $U_u{W}_{N{D}^2}=D{\gamma }_u{W}_N{U}_v$ has ${\gamma }_u=\left(\begin{array}{cc}(1+Nuv)/D& u\\ Nv& D\end{array}\right)\in {\Gamma }_0(N)$. A positive scalar matrix acts trivially. The nebentypus supplies $\psi (D)$; reindexing gives $\overline{\chi }(u)=\chi (-N)\chi (v)$. Thus the factor is initially $\psi (D)\chi (-N)\tau (\chi )/\tau (\overline{\chi })$. The primitive Gauss sum of a Dirichlet character identity $\tau (\chi )\tau (\overline{\chi })=\chi (-1)D$ gives the displayed equivalent factor. No assumption that $f$ itself is a Fricke eigenform is needed.