= Solution
A counterclockwise planar <rotation matrix> has the form
$$
R_\theta=
\begin{pmatrix}
\cos\theta&-\sin\theta\\
\sin\theta&\cos\theta
\end{pmatrix},
\qquad\theta\in\mathbb R.
$$
To recover this form from preservation of the <Euclidean norm>, let $\mathbf u,\mathbf v$ be the columns of $R$. Applying the norm condition to the two coordinate <unit vectors> gives $|\mathbf u|=|\mathbf v|=1$. Applying it to their sum gives
$$
2=|\mathbf u+\mathbf v|^2
=|\mathbf u|^2+2\mathbf u\cdot\mathbf v+|\mathbf v|^2,
$$
so $\mathbf u\cdot\mathbf v=0$. The columns are an <orthonormal basis>, and $R^TR=I$: $R$ is an <orthogonal matrix>.
Write $\mathbf u=(\cos\theta,\sin\theta)^T$. There are two perpendicular <unit vectors> available for $\mathbf v$, namely $\pm(-\sin\theta,\cos\theta)^T$. The first gives <determinant> $1$, the second $-1$. The positive determinant hypothesis selects the first. \b[A norm-preserving planar map with positive determinant is a rotation.]
For the commutation condition, write
$$
A=\begin{pmatrix}a&b\\c&d\end{pmatrix}.
$$
Direct <matrix multiplication> gives
$$
AJ=\begin{pmatrix}b&-a\\d&-c\end{pmatrix},
\qquad
JA=\begin{pmatrix}-c&-d\\a&b\end{pmatrix}.
$$
Equality forces $c=-b$ and $d=a$, so, with $u=a$ and $v=-b$,
$$
A=\begin{pmatrix}u&-v\\v&u\end{pmatrix}=uI+vJ.
$$
If $u=v=0$, take $\lambda=0$ and $R=I$. Otherwise set $\lambda=\sqrt{u^2+v^2}>0$ and choose $\cos\theta=u/\lambda$, $\sin\theta=v/\lambda$. Then
$$
\boxed{A=\lambda R_\theta,\qquad\lambda=\sqrt{u^2+v^2}.}
$$
This describes the <centralizer of a planar quarter-turn>. Its nonzero elements act as a rotation followed by a uniform scaling, just as multiplication by a nonzero <complex number> does.
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