= Solution
A trace-zero <Hermitian matrix> has the unique form
$$
\begin{pmatrix}h&w\\\overline w&-h\end{pmatrix},
\qquad h\in\mathbb R,\quad w\in\mathbb C.
$$
Set $x_3=h$, $x_1=\operatorname{Re}w$ and $x_2=-\operatorname{Im}w$. Then it is exactly $R(x)$; these assignments also prove uniqueness. This is the real <Pauli matrix> representation $R(x)=x_1\sigma_1+x_2\sigma_2+x_3\sigma_3$.
For a <unitary matrix>, $U^{-1}=U^\dagger$. Thus
$$
(U^{-1}AU)^\dagger=U^\dagger A^\dagger(U^{-1})^\dagger=U^{-1}AU,
$$
and invariance of the <matrix trace> under <similarity transformations> gives $\operatorname{tr}(U^{-1}AU)=0$. Therefore conjugation maps the space to itself.
Finally,
$$
\det R(x)=-x_3^2-(x_1-ix_2)(x_1+ix_2)=-|x|^2.
$$
The <determinant> is invariant under <similarity transformations>, so $\det R(y)=\det R(x)$ implies
$$
\boxed{|y|=|x|.}
$$
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