Combinatorial problem: Directed Acyclic Graph How many unique landscapes exist in 5D DAG (directed acyclic graph)? There are $2^5$ points (eg: 00000, 00001, ... 11111) and $(2^5)!$ combinations.
The problem is a combinatorial problem. It should be fun and interesting, and I am interested in discussing the solution here as well.
It is classifying all distinct 5 dimensional landscapes. A landscape in 5-space is an assignment of edge direction to each edge between vertices such that a directed acyclic graph is formed (DAG). 
Classification might include the number of peaks, basins, or a metric like that.
There are $(2^5)! $ combinations so obviously iterating through each combination and testing if it is a new landscape or an orientation of an old one won't work.
For example in the 2D case, there are $(2^2)!$ permutations  = $24$. This 24 is made up of 3 landscapes. There are 8 orientations of each one.
I am reluctant to draw a picture at first, because maybe the way you visualize it will help you find a solution.
 A: This is really only a partial answer. What I think you are asking for is a calculation of the number of non-isomorphic acyclic orientations of the hypercube $Q_5$ where two orientations are isomorphic if there is an automorphism of $Q_5$ taking one to the other.
This can be done as an (almost) straightforward application of Burnside's Lemma as described by Peter Cameron in these notes.
First, as Rebecca suggested, you should recall the pretty amazing result of Stanley that if $\chi_G(x)$ denotes the chromatic polynomial of a graph $G$ on $n$ vertices, then $(-1)^n\chi_G(-1)$ is equal to the number of acyclic orientations of $G$. 
Now, let $G$ be a graph and let $Aut(G)$ be its automorphism group. If $\sigma\in Aut(G)$, let $G/\sigma$ be the graph obtained by identifying the vertices in the orbits (i.e., cycles) of $\sigma$. This graph may contains loops and multiedges but for the statements below, multiedges may safely be replaced by a single edge.

Theorem: The number of non-isomorphic acyclic orientations that can be placed on $G$ is equal to $$\frac{1}{|Aut(G)|} \sum_{\sigma\in Aut(G)} (-1)^{sgn(\sigma)}\chi_{G/\sigma}(-1).$$

Note that if $G/\sigma$ has a loop, then $\chi_{G/\sigma}(x)=0$. 
As for the hypercube graph $Q_n$, $Aut(Q_n)$ is isomorphic to the hyperoctohedral group, so in principle the above formula can be evaluated by anyone determined enough to do so. But even for $n=5$, Maple has trouble finding the chromatic polynomial of $Q_5$ using the basic command, and I'm not expert enough to coax an answer from it. Perhaps someone else can make that calculation.
On the other hand, the case $n=2$ is easy enough that I will do it here to illustrate the theorem. In this case, if we label the vertices of $Q_2$ in cyclic order as $1,2,3,4$, then $$Aut(Q_2)=\{(1),(1\,2\,3\,4), (1\,3)(2\, 4), (1\,4\,3\,2),(2\,4),(1\,2)(3\,4),(1\,3),(1\,4)(2\,3)\}.$$
But we really only need consider $\sigma\in\{(1),(1\,3)(2\, 4), (2\,4),(1\,3)\}$ since the other $\sigma$ are such that $G/\sigma$ has loops.
For $\sigma=(1)$, $G/\sigma=G$. The chromatic polynomial of the $4$-cycle is $(x-1)^4+(x-1)$.
For $\sigma=(1\,3)(2\,4)$, $G/\sigma$ is isomorphic to a path on $2$ vertices and so $$\chi_{G/\sigma}(x)=x(x-1).$$
For $\sigma=(1\,3)$ or $(2\,4)$, $G/\sigma$ is isomorphic to a path on $3$ vertices and so $$\chi_{G/\sigma}(x)=x(x-1)^2.$$
Applying the theorem gives that the number of non-isomorphic acycle orientations of $Q_2$ is obtained by plugging in $x=-1$ into $$\frac{1}{8}\left( (x-1)^4+(x-1) + x(x-1)- x(x-1)^2 - x(x-1)^2\right),$$ which, as you already know, is equal to $3$.
Added: A lower bound for $Q_5$ can be obtained using results found in a paper of Kahale and Schulman. Their Theorem 6 shows that $$(-1)^n\chi_G(-1)\geq \prod_{v\in V(G)} (deg(v)+1)!^{\frac{1}{\deg(v)+1}}.$$
Since $Q_5$ has 32 vertices, is $5$-regular and $|Aut(Q_5)|= 5!(2^5)$, we see that there are at least $$\left\lceil\frac{6!^{\frac{32}{6}}}{5!(2^5)}\right\rceil = 451, 622, 346, 888 $$ non-isomorphic acyclic orientations. I have no idea how close that is to the exact number, but do you really need many more?
