I found the following result:
If $V$ is a euclidean vector space of finite dimension $n$ and $B$ is an orthonormal basis of $V$, then an endomorphism $f:V\to V$ is an isometry iff its matrix respect to $B$ is orthogonal.
The proof is not hard. If $f$ is an isometry and $A$ is its matrix respect to $B$, then \begin{equation*} X^\top \mathbb{I} Y =\langle x, y \rangle =\langle f(x),f(y)\rangle =(AX)^\top \mathbb{I} (AY)=X^\top A^\top A Y, \end{equation*} where $X$ and $Y$ are the coordinates of the vectors $x$ and $y$ with respect to $B$, respectively. As the vectors were arbitrarily chosen, it follows that $A^\top A=\mathbb{I}$, so $A$ is orthonormal. The reciprocal statement is obvious from this.
Then, I thought that the orthogonal group can be equal to the isometry group, but I read that this is not true: the isometry group has order $n(n+1)/2$ (because include the traslations) and the orthogonal group has order $n(n-1)/2$. However, I cannot see why is this way.
Translations aren't linear (they don't take $0$ to $0$); but are isometries. There are $n$ dimensions of translations in dimension $n$, accounting for the difference.