Here is a sketch of these proofs. The details of this are described in great detail in Michael Artin's Algebra - a book well-worth reading.
1a) For the $2\times 2$ case: If $A$ is orthogonal, then multiply by a suitable reflection to ensure that $\det(A) = 1$. Then note that
$$
A = \begin{pmatrix}
a & b \\
c & d
\end{pmatrix}
$$
where the vector $(a,c)$ and $(b,d)$ are orthonormal. Check that this must happen iff $\exists \theta \in [0,2\pi]$ such that
$$
a = \cos(\theta) = d, \text{ and } b = \sin(\theta) = -c
$$
1b) For the $3\times 3$: Again, assume $\det(A) = 1$. Now one needs to prove that $1$ is an eigen-value of $A$. This gives a one-dimensional subspace $W$ which is invariant under $A$. One can then prove that $W^{\perp}$ is also invariant under $A$, so $A$ will now restrict to a $2\times 2$ orthogonal matrix on $W^{\perp}$. Use part (a) to conclude that $A$ must be a rotation.
2) First note that if $T$ is an isometry, so is $T - T(0)$, so you may assume that $T(0) = 0$. Now since $\|T(x)\|^2 = \|x\|^2$ for all $x$, you can use the Polar identity to conclude that
$$
\langle Tx,Ty\rangle = \langle x,y\rangle \quad\forall x,y
$$
and this gives that $T$ must be an orthogonal matrix.
Here's a way to approach this problem from the linear algebra point of view. And in this answer I'm going to side-step the issue of orientation reversing isometries, which are not represented by Möbius transformation, instead by anti-Möbius transformations. So, I will explain how to use linear algebra to classify orientation preserving (o.p.) isometries, equivalently to classify Möbius transformations.
The classification of o.p. isometries is invariant under conjugacy. To see some examples of this from a synthetic point of view (i.e. no linear algebra), consider an o.p. isometry $\phi$, another o.p. isometry $\psi$, and the conjugate o.p. isometry $\psi \phi \psi^{-1}$.
First, if $\phi$ has a fixed point $x$ then $\psi \phi \psi^{-1}$ also has a fixed point, namely $\psi(x)$. Proof:
$$(\psi \phi \psi^{-1})(\psi(x)) = \psi \phi(x) = \psi(x)
$$
Furthermore, if $\phi$ rotates by an angle $\theta$ around $x$ then $\psi$ also rotates by an angle $\theta$ around $\phi(x)$. Proof: take a ray $R$ based at $x$, and the angle from $R$ to $\phi(R)$ at $x$ equals $\theta$; so the ray from $\psi(R)$ to $\psi(\phi(R))$ at $\psi(x)$ equals $\theta$.
Second, and with similar proofs, if $\phi$ fixes some geodesic line $L$ then $\psi \phi \psi^{-1}$ also fixes a line, namely $\psi(L)$. Furthermore, if $\phi$ translates a distance $d$ along $L$ then $\psi \phi \psi^{-1}$ translates a distance $d$ along $\psi(L)$.
Third, a parabolic transformation is one which fixes a single point on the circle at infinity and fixes no finite point.
Here's where things get interesting. Now you have to convince yourself of two further issues.
- Every o.p. isometry either fixes a point and rotates about it by some angle, or fixes some geodesic line and translates along that line by some distance, or is a parabolic isometry, or is the identity.
- Any two o.p. isometries which are rotations by the same angle, or are translations by the same distance, are conjugate. A similar statement holds for parabolic isometries, which I'll hold off until the end.
It is possible to prove this synthetically, but Möbius transformations certainly help. Here's an outline.
As usual we represent Möbius transformations by elements of $PSL(2,\mathbb{R}) = SL(2,\mathbb{R}) / \pm I$. Here's some algebra facts which you can easily verify.
- Given $M \in SL(2,\mathbb{R})$, $\text{trace}(M)$ is a conjugacy invariant in $SL(2,\mathbb{R})$. Also, its absolute trace $|\text{trace}(M)|$ is a conjugacy invariant in $PSL(2,\mathbb{R})$.
- If $|\text{trace}(M)|$ is not equal to 2 then it is a complete conjugacy invariant, i.e. if $M_1,M_2$ have the same absolute trace not equal to 2 then they are conjugate in $PSL(2,\mathbb{R})$. The case of absolute trace equal to $2$ corresponds to parabolic transformations and is not quite as clean; I'll skip that until the end.
Now, blend the geometry and the algebra: for each value of the absolute trace that is not equal to $2$, construct an example, verify that it is one of the isometries in the above list, and verify that every distinct isometry in the above list corresponds to exactly one value of the absolute trace. Having done that, the classification is complete! (Outside of the issue that I have ignored the parabolic case where absolute trace equals 2).
For example, absolute trace equal to zero is represented by $\pmatrix{0 & 1 \\ -1 & 0} : z \mapsto -\frac{1}{z}$ which is a rotation about the fixed point $0+1i$ in the upper half plane, with rotation angle $\pi$. More generally, absolute trace in the interval $[0,2)$ is always a rotation about a fixed point, with angles varying over $(0,\pi]$ by a simple strictly monotonic continuous function $[0,2) \mapsto (0,\pi]$ (whose formula some people remember but I never can).
For another example, absolute trace equal to $\frac{5}{2}$ is represented by $\pmatrix{2 & 0 \\ 0 & 1/2}$ which is a translation along the line $\text{RealPart}(z)=0$ by a distance $\ln(4)$. More generally, absolute trace in the inverval $(2,\infty)$ is always a translation along a line, with translation distance varying over $(0,\infty)$ by a simple strictly monotonic continuous function $(2,\infty) \mapsto (0,\infty)$ (again, with an explicit formula).
A few last words about absolute trace equal to $2$. Each of these represents either the identity or an o.p. parabolic isometry. In the full group of generalized Möbius transformations there is just one conjugacy class of parabolic isometry. But in $\text{PSL}(2,\mathbb{R})$ there are two conjugacy classes, one rotating "positively" around the fixed point at infinity and the other rotating "negatively".
Best Answer
$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Cpx}{\mathbf{C}}$When you speak without qualification of an isometry with respect to the standard Hermitian norm, you're speaking of an isometry of Euclidean $4$-space. The story of Euclidean isometries is more-or-less uniform independently of dimension, so let's speak of Euclidean $n$-space with $n$ an arbitrary positive integer. Here are some useful factlets:
If an isometry $F$ fixes the origin and the standard basis vectors, then $F$ is the identity.
Sketch of proof: This is clear if $n = 1$. To argue inductively on the dimension, decompose a point of $\Reals^{m+1}$ as $(x, x_{m}) = ((x_{1}, \dots, x_{m}), x_{m+1})$, and show $F$ preserves both $\|x\|$ (so $F(x) = x$ by the inductive hypothesis) and the $(m + 1)$th coordinate $x_{m+1}$.
If an isometry $F$ fixes the origin, there exists a unique $n \times n$ (real) orthogonal matrix $A$ such that $F(x) = Ax$ for all $x$ in $\Reals^{n}$.
Sketch: By the isometry condition and the polarization identity $u \cdot v = \frac{1}{4}(\|u + v\|^{2} - \|u - v\|^{2})$, the points $F(e_{i})$ constitute an orthonormal set in $\Reals^{n}$. Let $A$ be the unique orthogonal matrix satisfying $F(e_{i}) = Ae_{i}$ for each $i$, and apply the first item to $A^{-1} \circ F$. (This answers your first question affirmatively.)
If $F$ is an isometry, there exists a unique $n \times n$ orthogonal matrix and a unique $b$ in $\Reals^{n}$ such that $$ F(x) = Ax + b\quad\text{for all $x$ in $\Reals^{n}$.} $$ (You already have this proof. :)
If $p$ is a point of $\Reals^{n}$ and $u$ is a unit vector, then reflection in the hyperplane through $p$ orthogonal to $u$ is the mapping $$ R(x) = x - 2[(x - p) \cdot u] u, $$ namely, $x$ minus "twice the $u$-component of $x - p$". For example, if $u = e_{n}$ and $p$ is the origin, then $$ R(x_{1}, \dots, x_{n-1}, x_{n}) = (x_{1}, \dots, x_{n-1}, x_{n}) - 2x_{n} e_{n} = (x_{1}, \dots, x_{n-1}, -x_{n}). $$ The Euclidean group is generated by the set of reflections. In fact, if $F$ is a Euclidean isometry, there exist at most $(n + 1)$ reflections (not uniquely-defined!) whose composition is $F$.
A non-trivial translation is a composition of two reflections. (Why?)
A rotation is the composition of at most $n$ reflections, and is always the composition of an even number of reflections (because a reflection reverses orientation while a rotation preserves orientation). Thus, for example, every rotation of Euclidean $3$-space is a composition of at most two reflections, while a rotation of Euclidean $4$- or $5$-space is a composition of at most four reflections. (To prove this inductively, note that if $u$ and $v$ are arbitrary unit vectors, there exists a reflection exchanging $u$ and $v$.)
If, on the other hand, you want to speak only of Euclidean isometries of $\Cpx^{m} \simeq \Reals^{2m}$ for which the "rotation part" is complex-linear, you have the much smaller "holomorphic isometry group" corresponding to the unitary group $U(m) \subset SO(2m)$.
Thinking of $\Cpx^{2}$, the set of complex lines (a special class of real $2$-plane) is the well-known complex projective line, topologically a $2$-sphere, while the set of (oriented) real $2$-planes is topologically a product $S^{2} \times S^{2}$. This shows geometrically how small $U(2)$ is compared to $SO(4)$,