= Solution
Preserving the Euclidean <scalar product> means $(Ax)^T(Ay)=x^Ty$ for every pair of real <vectors> $x,y$. Since these bilinear expressions agree on all pairs of <basis> <vectors>, this is equivalent to $A^TA=I$: the columns of $A$ form an <orthonormal basis>.
The first column is a <unit vector>, so it is $(\cos\theta,\sin\theta)^T$ for some real $\theta$. A perpendicular <unit vector> is one of the two choices $\varepsilon(-\sin\theta,\cos\theta)^T$, where $\varepsilon\in\{1,-1\}$. Hence \b[all the required <orthogonal matrices> are]
$$
\boxed{A=R(\theta)D_\varepsilon=\begin{pmatrix}\cos\theta&-\varepsilon\sin\theta\\\sin\theta&\varepsilon\cos\theta\end{pmatrix},\quad D_\varepsilon=\operatorname{diag}(1,\varepsilon),\quad\theta\in\mathbb R.}
$$
Here $R(\theta)$ is a <planar rotation>, and $\theta$ is determined modulo $2\pi$. To check the <group> laws explicitly, $D_\varepsilon R(\phi)=R(\varepsilon\phi)D_\varepsilon$, so
$$
(R(\theta)D_\varepsilon)(R(\phi)D_\eta)=R(\theta+\varepsilon\phi)D_{\varepsilon\eta}.
$$
This proves closure. The identity is $R(0)D_1=I$, the inverse of $R(\theta)D_\varepsilon$ is $R(-\varepsilon\theta)D_\varepsilon$, and associativity is inherited from <matrix multiplication>. Thus the set is the <orthogonal group> $O(2)$.
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