Here is a more worked-out and concrete version of the proposed counterexample mentioned in the "Update" of the post.

1. We start from a regular arrangement of 12 great circles on the 3-sphere $\mathbb{S}^3$: 12 circles from the Hopf fibration, shown in stereographic projection to $\mathbb{R}^3$. They are the inverse image of the 12 corners of the regular icosahedron under the Hopf map $\mathbb{S}^3\to\mathbb{S}^2$.

2. On each circle we place 70 equidistant points: 840 points on 12 circles.

3. The supporting hyperplanes of the 3-sphere at these points form a 4-polytope $P$ with 840 equal facets: A perspective view of a facet, a 3-polytope with 40 vertices and 22 sides. (In order to ensure that all faces are equal, the regular 70-gons on the different circles cannot just be placed arbitrarily in Step 2. The points form the orbit of a specially chosen starting point under the group $\pm[I\times C_7]$.)

The facets are thin plates of roughly pentagonal shape centered at points of a circle and lying perpendicular to the circle. The plates stack up to form a twisted pentagonal tube that surrounds one circle. Such a tube makes one full turn of $360^\circ$ as it winds around the circle. The 12 tubes fill the space around the 3-sphere and enclose it completely.

The symmetry group $G$ of $P$ has 8400 elements: a given plate can be mapped to any of the 840 plates in 10 different ways. There are only rotations (determinant $+1$), no reflections (determinant $-1$). (Because of the special choice of the starting point of the orbit, the group $G$ is larger than the group $\pm[I\times C_7]$ by which the orbit was generated. I guess, with a generic orbit, the symmetry group will reduce to size 840, but the picture will be messier.)

The polytope $P$ is clearly not a uniform polytope: none of its 2-faces is a regular polygon. To definitely answer the question, one would have to argue why $G$ is not a subgroup of the symmetries of a different, uniform, polytope $P'$. One could use the classification of the point groups from the book of Conway and Smith, but maybe there is a more direct argument. The group must contain all rotations of a regular 70-gon.

This example and the programs (in Sage) that were used to produce the images were prepared with the help of my student Laith Rastanawi.

Here is a more worked-out and concrete version of the proposed counterexample mentioned in the "Update" of the post.

1. We start from a symmetric arrangement of 12 great circles $F_1,\ldots,F_{12}$ on the 3-sphere $\mathbb{S}^3$: 12 circles from the Hopf fibration, shown in stereographic projection to $\mathbb{R}^3$. They are the inverse image of the 12 corners of the regular icosahedron under the Hopf map $\mathbb{S}^3\to\mathbb{S}^2$ (fibers of the Hopf fibration).

2. On each circle we place 70 equidistant points: 840 points on 12 circles.

3. The supporting hyperplanes of the 3-sphere at these points form a 4-polytope $P$ with 840 equal facets: A perspective view of a facet, a 3-polytope with 40 vertices and 22 sides. (In order to ensure that all faces are equal, the regular 70-gons on the different circles cannot just be placed arbitrarily in Step 2. The points form the orbit of a specially chosen starting point under the group $\pm[I\times C_7]$.)

The facets are thin plates of roughly pentagonal shape centered at points of a circle $F_i$ and lying perpendicular to $F_i$. The plates stack up to form a twisted pentagonal tube that surrounds $F_i$. Such a tube makes one full turn of $360^\circ$ as it winds around the circle. The 12 tubes fill the space around the 3-sphere and enclose it completely.

The symmetry group $G$ of $P$ has 8400 elements: a given plate can be mapped to any of the 840 plates in 10 different ways. There are only rotations (determinant $+1$), no reflections (determinant $-1$). (Because of the special choice of the starting point of the orbit, the group $G$ is larger than the group $\pm[I\times C_7]$ by which the orbit was generated. I guess, with a generic orbit, the symmetry group will reduce to size 840, but the picture will be messier.)

The polytope $P$ is clearly not a uniform polytope: none of its 2-faces is a regular polygon. To definitely answer the question, one would have to argue why $G$ is not a subgroup of the symmetries of a different, uniform, polytope $P'$. One could use the classification of the point groups from the book of Conway and Smith, but maybe there is a more direct argument. The group must contain all rotations of a regular 70-gon.

This example and the programs (in Sage) that were used to produce the images were prepared with the help of my student Laith Rastanawi.

Here is a more worked-out and concrete version of the proposed counterexample mentioned in the "Update" of the post.

1. We start from a regular arrangement of 12 great circles on the 3-sphere $\mathbb{S}^3$: 12 circles from the Hopf fibration, shown in stereographic projection to $\mathbb{R}^3$. They are the inverse image of the 12 corners of the regular icosahedron under the Hopf map $\mathbb{S}^3\to\mathbb{S}^2$.

2. On each circle we place 70 equidistant points: 840 points on 12 circles.

3. The supporting hyperplanes of the 3-sphere at these points form a 4-polytope $P$ with 840 equal facets: A perspective view of a facet, a 3-polytope with 40 vertices and 22 sides. (In order to ensure that all faces are equal, the regular 70-gons on the different circles cannot just be placed arbitrarily in Step 2. The points form the orbit of a specially chosen starting point under the group $\pm[I\times C_7]$.)

The facets are thin plates centered at points of a circle and lying perpendicular to the circle. The plates along one circle stack up to form a twisted pentagonal tube. The tube makes one full turn of $360^\circ$ as it winds around the circle. The 12 tubes fill the space around the 3-sphere and enclose it completely.

The symmetry group $G$ of $P$ has 8400 elements: a given plate can be mapped to any of the 840 plates in 10 different ways. There are only rotations (determinant $+1$), no reflections (determinant $-1$). (Because of the special choice of the starting point of the orbit, the group $G$ is larger than the group $\pm[I\times C_7]$ by which the orbit was generated. I guess, with a generic orbit, the symmetry group will reduce to size 840, but the picture will be messier.)

The polytope $P$ is clearly not a uniform polytope: none of its 2-faces is a regular polygon. To definitely answer the question, one would have to argue why $G$ is not a subgroup of the symmetries of a different, uniform, polytope $P'$. One could use the classification of the point groups from the book of Conway and Smith, but maybe there is a more direct argument. The group must contain all rotations of a regular 70-gon.

This example and the programs (in Sage) that were used to produce the images were prepared with the help of my student Laith Rastanawi.

Here is a more worked-out and concrete version of the proposed counterexample mentioned in the "Update" of the post.

1. We start from a regular arrangement of 12 great circles on the 3-sphere $\mathbb{S}^3$: 12 circles from the Hopf fibration, shown in stereographic projection to $\mathbb{R}^3$. They are the inverse image of the 12 corners of the regular icosahedron under the Hopf map $\mathbb{S}^3\to\mathbb{S}^2$.

2. On each circle we place 70 equidistant points: 840 points on 12 circles.

3. The supporting hyperplanes of the 3-sphere at these points form a 4-polytope $P$ with 840 equal facets: A perspective view of a facet, a 3-polytope with 40 vertices and 22 sides. (In order to ensure that all faces are equal, the regular 70-gons on the different circles cannot just be placed arbitrarily in Step 2. The points form the orbit of a specially chosen starting point under the group $\pm[I\times C_7]$.)

The facets are thin plates of roughly pentagonal shape centered at points of a circle and lying perpendicular to the circle. The plates stack up to form a twisted pentagonal tube that surrounds one circle. Such a tube makes one full turn of $360^\circ$ as it winds around the circle. The 12 tubes fill the space around the 3-sphere and enclose it completely.

The symmetry group $G$ of $P$ has 8400 elements: a given plate can be mapped to any of the 840 plates in 10 different ways. There are only rotations (determinant $+1$), no reflections (determinant $-1$). (Because of the special choice of the starting point of the orbit, the group $G$ is larger than the group $\pm[I\times C_7]$ by which the orbit was generated. I guess, with a generic orbit, the symmetry group will reduce to size 840, but the picture will be messier.)

The polytope $P$ is clearly not a uniform polytope: none of its 2-faces is a regular polygon. To definitely answer the question, one would have to argue why $G$ is not a subgroup of the symmetries of a different, uniform, polytope $P'$. One could use the classification of the point groups from the book of Conway and Smith, but maybe there is a more direct argument. The group must contain all rotations of a regular 70-gon.

This example and the programs (in Sage) that were used to produce the images were prepared with the help of my student Laith Rastanawi.