Let $\mathbb{P}^2 = \mathbb{P}(\operatorname{Sym}^2\mathbb{C}^2)$ be the projective space of $2\times 2$ symmetric matrices over $\mathbb{C}$ modulo scalar.

Define an $\mathrm{SL}(2)$-action on $\mathbb{P}^2$ by $(A,Z)\mapsto AZA^t$, where $A\in \mathrm{SL}(2)$ and $Z\in \mathbb{P}^2$. The stabilizer of the identity is then
$$S = \{A\in \mathrm{SL}(2)\: | \: AA^t  = \lambda I\}$$
for some $\lambda\in\mathbb{C}^{*}$. However, some papers refer to $\mathbb{P}^2$ as the wonderful compactification of $\mathrm{SL}(2)/\mathrm{SO}(2)$ even though it seems to me that it would be more correct to say that $\mathbb{P}^2$ is the wonderful compactification of $\mathrm{SL}(2)/S$. 

For instance, via the map $\mathrm{SL}(2)\rightarrow\mathbb{P}^2,\: A\mapsto AA^t$ the matrix 
$$
\left(
\begin{array}{cc}
-i & 0 \\ 
0 & i
\end{array} 
\right)
$$
is mapped to 
$$
-\left(
\begin{array}{cc}
1 & 0 \\ 
0 & 1
\end{array} 
\right)
=
\left(
\begin{array}{cc}
1 & 0 \\ 
0 & 1
\end{array} 
\right)
$$
Note that 
$$
\left(
\begin{array}{cc}
-i & 0 \\ 
0 & i
\end{array} 
\right)\in S\setminus \mathrm{SO}(2)
$$
so the morphism $\mathrm{SL}(2)/\mathrm{SO}(2)\rightarrow\mathbb{P}^2$ is not of degree one. 

What am I misunderstanding here?