Is every finite group a group of "symmetries"?
Here's a sketch of an ugly argument. First construct an undirected graph whose automorphism group is G. You can do this by starting with a vertex vg for each element g of G and gluing in a path of length n(g-1h) from vg to vh with an extra leaf attached to the internal vertex of this path nearest to vg. Here n is an injective function from G to {3, 4, 5, ...}. (This construction might need |G| >= 4 or so, but clearly we can handle the smaller cases by hand.)
Now make the vertices of this graph into a metric space where the distance between two nonadjacent vertices is 1 and the distance between two adjacent vertices is 1-ε. Now we need to embed this metric space into some R^N. You can either appeal to some results about embedding finite metric spaces into R^N, or convince yourself that the map (x1, ..., xn) in (R^N)^n |-> (d(xi, xj)^2)_{1 <= i < j <= n} is a submersion near the regular simplex of edge length 1.
The permutohedron may have additional symmetries. For example, the order 3 permutohedron $\{(1,2,3),(1,3,2),(2,1,3),(3,1,2),(3,2,1)\}$ is a regular hexagon contained in the plane $x+y+z=6$, which has more than 6 symmetries.
I think we can solve it as follows:
Let $G$ be a group with finite order $n$ thought via Cayley's representation as a subgroup of $S_n$.
Let $S=\{A_1,...,A_n\}$ be the set of vertices of a regular simplex centered at the origin in an $(n-1)$-dimensional real inner product space $V$. Let $r$ be the distance between the origin and $A_1$. The set of vertices $S$ is an affine basis for $V$.
First unproven claim: If a closed ball that has radius $r$ contains $S$, then it is centered at the origin. Let $B$ be this ball.
The group of isometries that fix $S$ hence contains only isometries that fix the origin and permute the vertices, which can be identified with $S_n$ in the obvious way. The same is true if we replace $S$ by its convex hull.
Now $G$ can be thought of as a group containing some of the symmetries of $S$.
Let $C=k(A_1+2A_2+3A_3+...+nA_n)/(1+2+...+n)$, with $k$ a positive real that makes the distance between $C$ and the origin a number $r'$ slightly smaller than $r$.
Let $GC=\{g(C) : g \in G\}$. It has $n$ distinct points, as a consequence of $S$ being an affine basis of $V$.
Let $P$ be the convex hull of the points of $S \cup GC$.
Remark: A closed ball of radius $r$ contains $P$ iff it is $B$. The intersection of the border of $B$ and $P$ is $S$.
Second unproven claim: The extremal points of $P$ are the elements of $S \cup GC$.
Claim: $G$ is the group of symmetries of $P$.
If $g$ is in $G$, $g$ is a symmetry of $GC$ and of $S$, and it is therefore a symmetry of $P$.
If $T$ is a symmetry of $P$, then $T(P)=P$, and in particular, $T(P)$ is contained in $B$, and hence $T(0)=0$ (i.e. $T$ is also a symmetry of $B$). $T$ must also fix the intersection of $P$ and the border of $B$, so $T$ permutes the points of $S$, and it can be thought of as an element $s \in S_n$ sending $A_i$ to $A_s(i)$. And since $T$ fixes the set of extremal points of $P$, $T$ also permutes $GC$. Let's see that $s$ is in $G$.
Since $T(C)$ must be an element $g(C)$ of $GC$, we have $T(C)=g(C)$. But since $T$ is linear, $T(C/k)=g(C/k)$. Expanding,
$(A_{s(1)}+2A_{s(2)}+\ldots+nA_{s(n)}/(1+\ldots+n)=(A_{g(1)}+2A_{g(2)}+\ldots+nA_{g(n)})/(1+\ldots+n).$
For each $i \in \{1,...,n\}$ the coefficient that multiplyes $A_i$ is $s^{-1}(i)/(1+...+n)$ in the left hand side and $g^{-1}(i)/(1+...+n)$ in the right hand side. It follows that $s=g$.
I think that, taking $n$ into account, the ratio $r'/r$ can be set to substantiate the second unproven claim. The first unproven claim may be a consequence of Jung's inequality.
EDIT: With the previous argument, we can represent a finite group of order $n$ as the group of linear isometries of a certain polytope in an $(n-1)$-dimensional real inner product space.
Now, if a finite group $G$ of linear isometries of an $(n-1)$-dimensional inner product space $V$ is given, can we define a polytope that has $G$ as its group of symmetries? Yes. I'll give a somehow informal proof.
Let $G=\{g_1,...,g_m\}$. Let $A=\{a_1,...,a_n\}$ be the set of vertices of a regular $n$-simplex centered at the origin of $V$. Let $S$ be the sphere centered at the origin that contains $A$, and let $C$ be the closed ball. Notice that $C$ is the only minimum closed ball containing $A$.
(Remark: The set $A$ need not be a regular simplex. It may be any finite subset of $S$ that intersects all the possible hemispheres of $S$. $C$ will then still be only minimum closed ball containing it.)
Remark: An isometry of $V$ is linear iff it fixes the origin.
Before proceeding, we need to be sure that the $m$ copies of $A$ obtained by making $G$ act on it are disjoint. If that is not the case, our set $A$ is useless but we can find a linear isometry $T$ such that $TA$ does the job. We consider the set $M$ of all linear isometries with the usual operator metric, and look into it for an isometry $T$ such that for all $(g,a)$ and $(h,b)$ distinct elements of $GxA$ the equation $g(Ta)=h(Tb)$ does not hold. Because each of the $n\cdot m(n\cdot m-1)$ equations spoils a closed subset of $M$ with empty interior(*), most of the choices of $T$ will do.
Let $K=\{ga: g \in G, a \in A\}$. We know that it has $n\cdot m$ points, which are contained in the sphere $S$. Now let $e$ be a distance that is smaller than a quarter of any of the distances between different points of $K$. Now, around each vertex $a=a_i$ of $A$ make a drawing $D_i$. The drawing consists of a finite set of points of the sphere $S$, located near $a$ (at a distance smaller than $e$). One of the points must be $a$ itself, and the others (if any) should be apart from $a$ and very near each other, so that $a$ can be easily distinguished. Furthermore, for $i=1$ the drawing $D_i$ must have no symmetries, i.e, there must be no linear isometries fixing $D_1$ other than the identity. For other values of $i$, we set $D_i={a_i}$. The union $F$ of all the drawings contains $A$, but has no symmetries. Notice that each drawing has diameter less than $2\cdot e$.
Now let $G$ act on $F$ and let $Q$ be the union of the $m$ copies obtained. $Q$ is a union of $n\cdot m$ drawings. Points of different drawings are separated by a distance larger than $2\cdot e$. Hence the drawings can be identified as the maximal subsets of $Q$ having diameter less than $2\cdot e$. Also, the ball $C$ can be identified as the only sphere with radius $r$ containing $Q$. $S$ can be identified as the border of $C$.
Let's prove that the set of symmetries of $Q$ is $G$. It is obvious that each element of $G$ is a symmetry. Let $T$ be an isometry that fixes $Q$. It must fix $S$, so it must be linear. Also, it must permute the drawings. It must therefore send $D_1$ to some $gD_i$ with $g \in G$ and $1 \leq i \leq n$. But $i$ must be 1, because for other values of $i$, $gD_i$ is a singleton. So we have $TD_1=gD_1$. Since $D_1$ has no nontrivial symmetries, $T=g$.
We have constructed a finite set $Q$ with group of symmetries $G$. $Q$ is not a polytope, but its convex hull is a polytope, and $Q$ is the set of its extremal points.
(*) To show that for any $(g,a)$ and $(h,b)$ distinct elements of $G \times A$ the set of isometries $T$ satisfying equation $g(Ta)=h(Tb)$ has empty interior, we notice that if an isometry $T$ satisfies the equation, any isometry $T'$ with $T'a=Ta$ and $T'b\neq Tb$ must do (since $h$ is injective). Such $T'$ may be found very near $T$, provided $\dim V>2$. The proof doesn't work for $n=1$ or $n=2$, but these are just the easy cases.
I think I have a solution that will work, but I'm not 100% certain.
Let V be a faithful representation of G (so the map G→GL(V) has no kernel). Pick a "sufficiently generic" point of V and consider the convex hull of the orbit of that point. G includes into the group of symmetries of this polytope, but it could have additional symmetries. For example, Z/4 acts on the plane by rotation. If you take any point, its orbit is a square, which has additional symmetries (reflections).
You can fix this by modifying the above construction as follows. Replace the point by a set of points S which is totally asymmetric (has no symmetries at all in GL(V)). Think of this set S as being a small cluster of points very far from the origin, so all the images of S under the action of G are easily distinguishable. Since S was totally asymmetric, the only symmetries of the union of these images should be elements of G.
If you're careful with how you chose your S, you shouldn't "lose too much asymmetry" when you take the convex hull of the union of the images of S.