= Solution
Let $T$ be an <isometry>, and set $S(x)=T(x)-T(0)$. Then $S(0)=0$, and preservation of distances gives $|S(x)|=|x|$. The <polarization identity> yields
$$
\langle S(x),S(y)\rangle=\frac12\bigl(|S(x)|^2+|S(y)|^2-|S(x)-S(y)|^2\bigr)=\langle x,y\rangle.
$$
For a standard <orthonormal basis> $e_1,e_2,e_3$, put $v_i=S(e_i)$. These three vectors are an <orthonormal basis>. Since $\langle S(x),v_i\rangle=x_i$, we have $S(x)=\sum_i x_iv_i=Qx$ for an <orthogonal matrix> $Q$. Thus
$$
\boxed{T(x)=Qx+b,\qquad Q^TQ=I,\quad b=T(0),}
$$
an <affine map>.
For a finite <isometry group> $G$, take $c=|G|^{-1}\sum_{g\in G}g(0)$. An affine map preserves an affine average, so for any $h\in G$,
$$
h(c)=\frac1{|G|}\sum_{g\in G}h(g(0))=\frac1{|G|}\sum_{g\in G}g(0)=c.
$$
\b[The orbit barycentre is a common <fixed point>.] This proves the <fixed point of a finite Euclidean isometry group> result.
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