= Solution
For an active clockwise <rotation> of column <vectors> in the usual oriented $xy$ plane, take
$$
R_{\rm cw}(\theta)=\begin{pmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{pmatrix}.
$$
Direct <matrix multiplication> gives
$$
R_{\rm cw}^TR_{\rm cw}
=\begin{pmatrix}\cos^2\theta+\sin^2\theta&0\\0&\cos^2\theta+\sin^2\theta\end{pmatrix}
=I.
$$
Thus the <rotation matrix> is an <orthogonal matrix>, preserving the Euclidean <inner product>, with <determinant> $+1$.
For a passive one-dimensional <Lorentz boost>, use <rapidity> $\eta$, $\beta=\tanh\eta$:
$$
\Lambda(\eta)=\begin{pmatrix}\cosh\eta&-\sinh\eta\\-\sinh\eta&\cosh\eta\end{pmatrix},\qquad
\Lambda^T\Lambda=\begin{pmatrix}\cosh2\eta&-\sinh2\eta\\-\sinh2\eta&\cosh2\eta\end{pmatrix}.
$$
This is not $I$ unless $\eta=0$. Hence a nontrivial <Lorentz boost> is not an <orthogonal matrix> for the Euclidean <inner product>. Instead it preserves the <Minkowski metric>:
$$
\boxed{\Lambda^Tg\Lambda=g,\qquad g=\operatorname{diag}(-1,1),}
$$
by $\cosh^2\eta-\sinh^2\eta=1$. Thus it is pseudo-orthogonal. The zero-<rapidity> identity is the explicit exception to the assertion that a <Lorentz boost> is not orthogonal.
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