An orientation-reversing symmetry forces $|\mathrm{Sym}^+(C)|\le d$

Identify the Euclidean plane with $\mathbb C$ and use the complex coordinates $X=z$, $Y=\bar z$. Let $P\in\mathbb C[X,Y]$ be irreducible in $\mathbb C[X,Y]$, of total degree $d=\deg P\ge 2$, and assume that its real locus $C=\{z\in\mathbb C: P(z,\bar z)=0\}$ is infinite and is not a circle: there are no $c\in\mathbb C$ and real $R>0$ with $C=\{z\in\mathbb C: |z-c|=R\}$. Let $\mathrm{Sym}^+(C)$ be the set of orientation-preserving isometries $T(z)=az+b$ ($a,b\in\mathbb C$, $|a|=1$) such that $T(z)\in C\iff z\in C$ for all $z\in\mathbb C$. Suppose that $C$ has an orientation-reversing symmetry, that is, there are $a_0,b_0\in\mathbb C$ with $|a_0|=1$ and $a_0\bar z+b_0\in C\iff z\in C$ for all $z\in\mathbb C$. Then

In the proof of Theorem 1 this is the case in which $C$ has a reflection; it gives $N\le d$ and hence $|\mathrm{Sym}(C)|\le 2d$. In the formalization it is also used to exclude orientation-reversing symmetries of the curves $C_{m,\alpha}$ of equation (1) with $m\ge 3$, and to count the symmetries of the curves $\operatorname{Re}(z^d)=1$.

Formalization Note: isometries are recorded by their pairs $(a,b)\in\mathbb C\times\mathbb C$, and the orientation-reversing symmetry is given as such a pair. The cardinality is the natural-number cardinality, which is $0$ for an infinite set; the finiteness of $\mathrm{Sym}^+(C)$ is a separate result.