= Solution
Multiplying numerator and denominator by the same nonzero constant does not change a <Möbius transformation>. Since $b\ne0$, divide by $b$ and relabel the coefficients so that $b=1$. On $|z|=1$, preservation of the circle gives
$$
|a+z|^2=|c+dz|^2.
$$
Equating the constant and first <Fourier coefficients> yields
$$
|a|^2+1=|c|^2+|d|^2,\qquad \bar a=\bar c\,d.
$$
If $c=0$, the second equality forces $a=0$, contradicting the nonzero <determinant>. Hence $d=\bar a/\bar c$. With $C=|c|^2>0$, the first equality becomes
$$
C^2-(1+|a|^2)C+|a|^2=0,
\qquad (C-1)(C-|a|^2)=0.
$$
The root $C=|a|^2$ would imply $ad=|a|^2/\bar c=c$, making the numerator and denominator proportional. It is excluded by nondegeneracy. Thus $|c|=1$ and $d=c\bar a$. Writing $1/c=e^{i\psi}$ gives
$$
\boxed{f(z)=e^{i\psi}\frac{a+z}{1+\bar a z},\qquad |a|\ne1.}
$$
Here $a$ is the normalized coefficient after division by the original $b$. The inequality $|a|\ne1$ is again the <determinant> condition. For $|a|<1$ the map preserves the disc; for $|a|>1$ it interchanges its inside and outside. Circle preservation alone does not justify restricting to <automorphisms of the unit disk>.
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