How many ways can you inscribe five 24-cells in a 600-cell, hitting all its vertices? You can inscribe five tetrahedra in a dodecahedron so that each vertex of the dodecahedron is the vertex of just one tetrahedron, as drawn here by Greg Egan:

Warmup question: How many ways can you do this?
I believe there are just two: the way shown here, and its mirror image.
But here's my real question.  I just noticed that the above phenomenon has a four-dimensional analogue.  The 24-cell and 600-cell are four-dimensional regular polytopes, and you can can inscribe five 24-cells in a 600-cell so that each vertex of the 600-cell is the vertex of just one 24-cell.  
Question: How many ways can you do this?
To see that it's possible at all, note that the unit quaternions form a group that is the double cover of $\mathrm{SO}(3)$.  Thus, the rotational symmetry group of the dodecahedron, $\mathrm{A}_5$, has a 120-element double cover that is a subgroup of the unit quaternions.  This is called the 'binary icosahedral group' $2\mathrm{I}$.  Its points, thought of as quaternions, are the vertices of the 600-cell, as drawn here by Robert Webb's Stella software:

Similarly, the double cover of the tetrahedron's rotational symmetry group $\mathrm{A}_4$ is called the 'binary tetrahedral group' $2\mathrm{T}$, and its points are the vertices of the 24-cell:

Since $\mathrm{A}_4 \subset \mathrm{A}_5$ is a subgroup with index 5, so is $2\mathrm{T} \subset 2\mathrm{I}$.  Thus, the five left cosets of $2\mathrm{T}$ in $2\mathrm{I}$ give a way of inscribing five 24-cells in a 600-cell so that each vertex of the 600-cell is the vertex of just one 24-cell.  
The right cosets give another method.   Furthermore, there are five ways to embed $\mathrm{A}_4$ as a subgroup of $\mathrm{A}_5$, which are actually the symmetry groups of the five tetrahedra in the dodecahedron in Egan's picture above.   So, we get 2 × 5 = 10 ways to inscribe five 24-cells in a 600-cell so that each vertex of the 600-cell is the vertex of just one 24-cell.  I believe, but haven't carefully checked, that these are all distinct.  I can't think of any other ways to do it.  Thus, I conjecture that the answer to my question is: 10.
 A: The 600-cell can be tiled by five 24-cells in exactly ten different ways. These are written explicitly in table 2 of "Parity proofs of the Bell-Kochen-Specker theorem based on the 600-cell", where you can also see an application of this fact to giving a proof of the Kochen-Specker theorem, ruling out the existence of noncontextual hidden variable theories in quantum mechanics.
The authors say that these ten different tilings were first discovered by P.H. Schoute and give a reference to Coxeter's book on regular polytopes.
A: $\newcommand\Z{\mathbb{Z}}$
This is an elementary answer to the warmup question.
There are precisely $10$ ways of inscribing a tetrahedron inside a dodecahedron. The symmetry group $G \subset O(3)$ of the dodecahedron acts on these tetrahedra. Note that $G = A_5 \times \Z/2\Z$. The stabilizer of a tetrahedron has to act by symmetries on the tetrahedron and thus be a subgroup of $S_4 \subset O(3)$ which acts on $T$. But $S_4$ is not a subgroup of $A_5$, and so the stabilizer (which must have order at least $10$ by the Orbit-Stabilizer theorem) is $A_4$, and the orbit of $G$ has size $10$, and $G$ is transitive.
On the other hand, since $A_4$ is contained in $A_5$, again by the Orbit-Stabilizer theorem, the action of $A_5$ is not transitive, and the $10$ tetrahedra are thus naturally divided up into two sets $U$ and $V$ of size $5$. 
Let $X$ denote the set of subsets of $U \cup V$ of order $5$. Then $|X| = \binom{10}{5} = 252$.
There is certainly an action of $G$ on $X$, and both $U$ and $V$ are the unique fixed points of this action.
We may think of an element $x \in X$ as giving a configuration of $5$ tetrahedra inside $D$, the question is which elements of $X$ hit every vertex of $D$.
Note that $U$ and $V$ give the two configurations indicated in your answer (which are invariant under the action of $A_5$). We seek the others (if they exist).
Lemma: If $T$ and $T'$ are two tetrahedra with no vertex in common, then the $8$ vertices $S$ of $T$ and $T'$ determine both $T$ and $T'$.
Proof: Certainly if we can determine one $T$ from $S$, we can determine $T'$ as well. Take a vertex $v \in S$. It lies on some $T$. The three other vertices of $T$ lie on a plane $P(v)$ corresponding (uniquely) to $v$. The plane $P$ has $6$ vertices which are divided up into two sets of $3$. In particular, if we cannot determine $T$ from $S$ and $v$, then $S$ must contain all $6$ verticies. But now choose a different point $w$ in $S$, and we determine that $P(w)$ also contains at least $6$ points of $S$, and hence $P(v) \cap P(w)$ contains at least $4$ points of $S$.  But the intersection of any two such planes is a line, and no four points of $D$ lie on a line.
Now we show the only $x \in X$ which hit every point on $P$ are $U$ and $V$. WLOG (by symmetry), we may assume that $x$ has at least $3$ tetrahedra in $U$. Let $T$ and $T'$ determine the two remaining tetrahedra in $x$ (which may or may not be in $U$). Since the complete set of vertices of $U$ is all of $D$, it follows that the two remaining tetrahedra in $U$ have the same vertices as $T$ and $T'$. Hence, by the Lemma, we must have $x = U$.
A: "The Geometry of $H_4$ Polytopes" by Denney et al. (pages 6-8) shows the answer to Puzzle 1 is no. Each hypercube within a 600-cell appears in just one 24-cell, and any two 24-cells are either disjoint or intersect in a hexagon, where each of the three orthoplexes in a 24-cell contains two of the hexagon's six vertices. Thus each of the three hypercubes in a 24-cell, consisting of two orthoplexes, contains four of the hexagon's six vertices; but since the seven hypercubes are disjoint (by assumption) there cannot be any hexagonal intersections, and so the seven 24-cells that contain them would have to be disjoint as well, which is clearly impossible.
