For an isometry $f:\mathbb{C}\rightarrow\mathbb{C}$, let Fix($f$) denote the set of fixed points of $f$. If $f:\mathbb{C}\rightarrow\mathbb{C}$ and $h:\mathbb{C}\rightarrow\mathbb{C}$ are isometries, show that Fix($h \circ f \circ h^{-1}) = h($Fix($f$)).
I know that $\rho_\theta(h(z)) = h(\rho_\theta,z_0(z))$ but I do not know if that even applies here or not.
Let $h : A \to B$ be a bijection between arbitrary sets $A, B$ and $f : A \to A$ be any function. Then $\text{Fix}(h \circ f \circ h^{-1}) = h(\text{Fix}(f))$.
To see this let $a \in \text{Fix}(f)$, i.e. $f(a) = a$. Then $(h \circ f \circ h^{-1})(h(a)) = h(f(h^{-1}(h(a)))) = h(f(a)) = h(a)$, i.e. $h(a) \in \text{Fix}(h \circ f \circ h^{-1})$. Conversely, let $b \in \text{Fix}(h \circ f \circ h^{-1})$, i.e. $(h \circ f \circ h^{-1})(b) = b$. Then $f(h^{-1}(b)) = h^{-1}(h(f(h^{-1}(b)))) = h^{-1}(b)$. This shows that $h^{-1}(b) \in \text{Fix}(f)$ and thus $b \in h(\text{Fix}(f))$.