= Solution
Let $b=f(0)$ and $g(x)=f(x)-b$. Distance preservation and polarization give $\langle g(x),g(y)\rangle=\langle x,y\rangle$. The vectors $g(e_1),g(e_2),g(e_3)$ form an orthonormal <basis>, and $\langle g(x),g(e_i)\rangle=x_i$. Hence $g(x)=\sum_i x_i g(e_i)=Qx$ for an orthogonal <matrix> $Q$. Every <Euclidean isometry> is therefore $f(x)=Qx+b$.
To decompose $Q$, if $Qe_1\ne e_1$ reflect in the origin plane normal to $Qe_1-e_1$. This reflection sends $Qe_1$ to $e_1$. The remaining orthogonal transformation fixes $e_1$ and acts on its two-dimensional orthogonal complement; repeat with $e_2$, and then on the last one-dimensional complement. At most three plane <reflections> are needed. A translation $x\mapsto x+b$ is itself a product of two reflections in parallel planes: for $\mathbf n=b/|b|$, reflecting first in $\mathbf n\cdot x=0$ and then in $\mathbf n\cdot x=|b|/2$ gives that translation. Thus every isometry is a finite composition of plane reflections, the <finite reflection decomposition of a Euclidean isometry>.
For an isometry fixing zero, the preceding construction gives $N\le3$. A plane reflection's linear part has $\operatorname{rank}(I-R)=1$. The identity $I-AB=(I-A)+A(I-B)$ implies that a product of $m$ such linear parts has rank of $I-Q$ at most $m$, even if the reflecting planes are affine. For central inversion $Q=-I$, this rank is three. The <reflection length of a Euclidean orthogonal map> therefore gives
$$
\boxed{N=3,\qquad x\mapsto-x\text{ requires three reflections}}.
$$
For the regular tetrahedron, center it at zero. Every permutation of its four vertices determines an isometry, and the orientation-reversing ones correspond to odd permutations in $S_4$: six transpositions and six four-cycles. Each transposition is a plane reflection fixing the two unswapped vertices; its plane contains their edge and the midpoint of the opposite edge. These give six distinct mirror planes.
The other six are rotoreflections: rotate by $+\pi/2$ or $-\pi/2$ about an axis through the midpoints of a pair of opposite edges, and then reflect in the plane through the center perpendicular to that axis. There are three axes and two senses. For vertices $(1,1,1),(1,-1,-1),(-1,1,-1),(-1,-1,1)$, the map $(x,y,z)\mapsto(-y,x,-z)$ is one such four-cycle; the other axes and inverse rotations give all six. Hence the complete orientation-reversing <tetrahedral symmetry> list is \b[six mirror reflections and six quarter-turn rotoreflections].
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