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The assertion is true. A Möbius map fixes infinity if and only if $c=0$: if $c\ne 0$, its value there is the finite number $a/c$. For $c=0$, the decomposition above uses only a nonzero complex scaling and a translation. These maps form the Affine subgroup of the Möbius group.

True. Conjugate the unique fixed point to infinity. The resulting map lies in the Affine subgroup of the Möbius group, so it has the form $g(z)=az+b$. A unique fixed point forces $a=1$ and $b\ne 0$; otherwise $g$ would have a finite fixed point as well. Conjugation by the scaling $z\mapsto z/b$ changes $z\mapsto z+b$ into $z\mapsto z+1$. Thus a nonidentity Möbius transformation with one fixed point is conjugate to a translation.

A Möbius transformation is generated by translations $z\mapsto z+b$, nonzero scalings $z\mapsto az$, and inversion $z\mapsto 1/z$. Translations and scalings plainly take circles and lines to circles and lines. More uniformly, every circle or line has an equationwhere $\alpha ,\gamma \in \mathbb{R}$; the case $\alpha =0$ is a line. Under inversion, put $z=1/w$ and multiply by $|w{|}^2$ to obtainwhich is again such an equation. The generators, and hence every Möbius transformation, therefore preserve Generalized circle under a Möbius transformation.

Now letIf $c\ne 0$, its finite pole $-d/c$ lies on some Euclidean circle, and that circle maps to a line because its image contains infinity. Thus $f$ takes every circle to a circle exactly when $c=0$, so the subgroup in question is the Affine subgroup of the Möbius group, consisting of $z\mapsto az+b$ with $a\ne 0$.

It is not normal. If $I(z)=1/z$ and $T_b(z)=z+b$ with $b\ne 0$, then $T_b$ is affine buthas a finite pole and is not affine. Hence conjugation by $I$ does not preserve the subgroup.