Normalizer of a subgroup of $GL_2(\mathbb{R})$ I have following subgroup of $GL_2(\mathbb{R}):$ $$A=\Bigg\{\left( \begin{array}{cc}
1 & 0  \\
0 & 1  \end{array} \right),\left( \begin{array}{cc}
0 & -1  \\
1 &  0  \end{array} \right),\left( \begin{array}{cc}
 0 & 1  \\
-1 & 0  \end{array} \right),\left( \begin{array}{cc}
-1 & 0  \\
0 & -1  \end{array} \right)\Bigg\}.$$
It's basically the rotation group of a square centered at the origin in $\mathbb{R}^2.$ 
I need to compute the normalizer of $A$. 
Let's denote the normalizer by $A_N.$ Then, by definition, $$A_N:=\{g\in GL_2(\mathbb{R}):  gA=Ag\}.$$ $$\equiv \{g\in GL_2(\mathbb{R}):  gag^{-1}\in A,\ \forall \ a\in A\}.$$ 
I am using the second definition to compute all such $g$ by hand, but the computation is leading me nowhere. I have a feeling that all matrices that have $1$ or $-1$  on one diagonal and $0$ in the other diagonal satisfy the normalization condition. For example: all elements of $A$ are clearly in $A_N$. When I check for other matrices with $1$ or $-1$ on one diagonal and $0$ in the other, they all satisfy the condition. How to make this idea more precise? I am sure there must be a nice way to arrive at such $g$ through computation but I am yet to find one.

$$\begin{pmatrix}a&b\\c&d\end{pmatrix}\in N_{GL_2(\Bbb R)}(A)\implies $$
$$\begin{align*}
\bullet&\begin{pmatrix}b&\!\!-a\\d&\!\!-c\end{pmatrix}=\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}0&\!\!-1\\1&0\end{pmatrix}=\begin{pmatrix}0&\!\!-1\\1&0\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}\!\!-c&\!\!-d\\a&b\end{pmatrix}\iff\begin{cases}a=d\\{}\\b=-c\end{cases}\\{}\\\bullet&\begin{pmatrix}\!\!-b&a\\\!\!-d&c\end{pmatrix}=\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}0&1\\\!\!-1&0\end{pmatrix}=\begin{pmatrix}0&1\\\!\!-1&0\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}c&d\\\!\!-a&\!\!-b\end{pmatrix}\iff\begin{cases}a=d\\{}\\b=-c\end{cases}\end{align*}$$
The other two cases are useless as those are scalar matrices and thus they commute with everything...
We thus get that the general form of an element of the normalizer is
$$\begin{pmatrix}a&b\\\!\!-b&a\end{pmatrix}\;,\;\;\text{and of course}\;\;a^2+b^2\neq0$$
