Finding determinant of matrix lie group

Question

Given $G=\left\{A\in M_2(\mathbb{R})\mid A^\top XA = X\right\}$ where $X = \pmatrix{3&1\\1&1}$ and a Lie algebra $\mathfrak g=\left\{Y\in M_2(\mathbb{R})\mid Y^\top X+XY = 0\right\}$, how would I show that every element in $G$ does not have $\det=-1$?

Answer 1

The matrix $$ A = \pmatrix{0&\sqrt{3}\\\frac{1}{\sqrt3}&0} $$ has $\det{A} = - 1$ and satisfies $A^T X A = X$ unless I miscalculated....

(The Group in question can be thought of as the orthogonal group relative to the scalar product induced by $X$, which has two components (in the sense of topology), like the usual $O(2)$. These are precisely the counterimages of $1, -1$ respectively, under $\det$)