Simple proof that symmetries of regular polyhedron fix its center? Let $P$ be some regular polyhedron in $\mathbb{R}^3$ (i.e. a regular $n$-hedron with $n = 4, 6, 8, 12,$ or $20$), centered at the origin $o = (0, 0, 0)$, and with vertices $v_1, ..., v_n$ all lying on the unit sphere $S^2$.
Now, suppose that $\varphi$ is an isometry1 on $\mathbb{R}^3$ with the property that, for some permutation $\sigma$ of $\{1,\dots,n\}$, we have $\varphi(v_i) = v_{\sigma(i)}, \; \forall \; i \in \{1,\dots,n\}$.
The problem is to show that such an isometry $\varphi$ must fix the origin $o$; i.e., that $\varphi(o) = o$.
I realize that this assertion is "intuitively obvious", but is there a simple proof of it that does not resort to hand-waving and appeals to geometric intuition?
I can come up only with rather laborious proofs.
Feel free to use the fact that any isometry can be "factored" into an isometric linear map followed by a translation.  In particular, feel free to assume the existence of an isometric linear map $\rho:\mathbb{R}^3 \to \mathbb{R}^3$ and a translation $\tau_b: v \mapsto v + b$ (for some vector $b \in \mathbb{R}^3$) such that $\varphi = \tau_b \circ \rho$.  With this assumption, the problem is equivalent to showing that $\tau_b(o) = o$.
1 By isometry I mean a map $f:\mathbb{R}^d \to \mathbb{R}^d$ that preserves the Euclidean metric $m$: i.e. $m(v, w) = m(f(v), f(w))$ for all $v, w \in \mathbb{R}^d$.
 A: Each of the regular polyhedra except for the tetrahedron contains a pair of opposite vertices, so that the center is the midpoint of the interval joining them.  Any isometry, being a composition of a linear map and translation, must preserve pairs of opposite points and therefore must fix the origin.
For the regular tetrahedron, the sum of the vectors defined by the four vertices is the origin.  Since an isometric linear map preserves the average of the four vertices, it must also fix the origin (and similarly for a translation).
Actually the latter argument also applies to the other regular polyhedra, since the average of the vertices is always 0.
A: The essential point is that any isometry $T:\>{\mathbb R}^d\to{\mathbb R}^d$ is an affine linear map
$$T:\quad x\mapsto Ax +b\ ,\tag{1}$$
where $A:\>{\mathbb R}^d\to{\mathbb R}^d$ is linear and $b\in{\mathbb R}^d$. It follows that your conjecture is an immediate consequence of the following
Lemma. When an affine linear map $(1)$ permutes $n\geq1$ given points $x_1$, $\ldots$, $x_n$ then their centroid
$c:={1\over n}\sum_{k=1}^n x_k$ stays fixed.
Proof. 
$$T(c)=Ac+b={1\over n}\sum_{k=1}^n Ax_k \ + \ b={1\over n}\sum_{k=1}^n\bigl( Ax_k + b\bigr)={1\over n}\sum_{k=1}^n T(x_k)={1\over n}\sum_{k=1}^n x_{\pi(k)}=c\ .$$
A: OK, I thought of a proof of a slightly stronger statement:
Proposition: Suppose that the points in the set $V = \{v_1, v_2, v_3, v_4\}$ lie on the sphere $S^2 \subset \mathbb{R}^3$, and are not coplanar.  Then, if $\varphi:\mathbb{R}^3\to\mathbb{R}^3$ is an isometry that fixes $V$ (i.e. $\varphi(V) = V$), it follows that $\varphi(o) = o = (0, 0, 0)$.

This proposition is a corollary of the theorem that says that, given any four non-coplanar points in $\mathbb{R}^3$, there exists a unique sphere that contains all four points1.
Proof: the fact that $\varphi$ is an isometry implies that (1) $\varphi(V)$ is contained in the sphere $\varphi(S^2)$ centered at $\varphi(o)$ (since $V \subset S^2$), and (2) the points in $\varphi(V)$ are not coplanar2.  It follows (from the theorem cited in the preceding paragraph) that $\varphi(S^2)$ is the only sphere that contains $\varphi(V)$.  But, by assumption, $\varphi(V) \subset S^2$.  Therefore it follows that $\varphi(S^2) = S^2$, and thus $\varphi(o) = o$.
1 This is the 3-dimensional analogue of the theorem that states that, given three non-collinear points, there exists a unique circle that contains all three.  These theorems are proved by showing that the locus of the centers of all the possible circles/spheres containing all the points in question consists of exactly one point.
2 My reasoning goes like this: a general isometry $\varphi$ is a composition of a linear isometry with a translation, both of which send lines to lines, and therefore, planes to planes.  Therefore, these transformations preserve coplanarity.  Furthermore, they are invertible, so they also preserve "non-coplanarity".  I wish I had a more elegant argument.
