For $ P(\mathbf{x}) = \mathbf{x} \cdot \mathbf{x} - 1 $ proof algebraically that $ \mathbf{Sim}_{\mathcal{A}}(P)=\mathbf{O}(2) $
Where, $ \mathbf{Sim}_{\mathcal{A}}(P):=\{g \in \mathcal{A}(2) \mid P \circ g=P\} $
My attempt
$ \subseteq$) Since $ \mathbf{Sim}_{\mathcal{A}}(P):=\{g \in \mathcal{A}(2) \mid P \circ g=P\} $, note that
$ \begin{align*} P \circ g & = P \\ (P \circ g)(\mathbf{x}) & = P(\mathbf{x}) \\ (P(g(\mathbf{x})) & = \mathbf{x} \cdot \mathbf{x} - 1 \end{align*} $
Thus, observe that $$ P(\mathbf{x}) \cdot P(\mathbf{x}) = (\mathbf{x} \cdot \mathbf{x} - 1) \cdot (\mathbf{x} \cdot \mathbf{x} - 1) $$ $$ P(g(\mathbf{x})) \cdot P(g(\mathbf{x})) = (\mathbf{x} \cdot \mathbf{x} - 1) \cdot (\mathbf{x} \cdot \mathbf{x} - 1)$$
It follows
$ \begin{align*} P(\mathbf{x}) \cdot P(\mathbf{x}) & = P(g(\mathbf{x})) \cdot P(g(\mathbf{x}))\\ % \mathbf{x} \cdot \mathbf{x} & = g(\mathbf{x}) \cdot g(\mathbf{x}) \end{align*} $
$$ \mathbf{Sim}_{\mathcal{A}}(P) \subseteq \mathbf{O}(2) $$
Not sure if that last step is correct (applied the inverse on both sides and distributed it)
$ \supseteq $) Suppose that $ g \in \mathbf{O}(2) $. Henceforth
$ \begin{align*} (P \circ g)(\mathbf{x}) \cdot (P \circ g)(\mathbf{y}) & = P(g(\mathbf{x})) \cdot P(g(\mathbf{y}))\\ & = P(g(\mathbf{x}) \cdot (g(\mathbf{y}))\\ & = P(\mathbf{x} \cdot \mathbf{y})\\ & = (\mathbf{x} \cdot \mathbf{x} -1) \cdot (\mathbf{y} \cdot \mathbf{y} -1) \end{align*} $
$$ \boxed{(P \circ g)(\mathbf{x}) \cdot (P \circ g)(\mathbf{y}) = P(\mathbf{x}) \cdot P(\mathbf{y})} $$
Since $ g \in \mathbf{O}(2) $, thus
$$ \mathbf{O}(2) \subseteq \mathbf{Sim}_{\mathcal{A}}(P) $$
The proof doesn't seem the be correct when applying the inverse in the first part and for the second part I have my doubts if the orthogonality of $g$ is being applied correctly. Moreover, in part two I'm using the fact that $ P(g(\mathbf{x})) \cdot P(g(\mathbf{y})) = P(g(\mathbf{x}) \cdot (g(\mathbf{y})) $ and I'd believe that I'd have to prove that first to use it (and maybe, it is not true!)