No. We assume that these numbers are real (if they are complex or quaternion or some type of vector over reals or something just split them up by a basis and then apply this argument). Take the sum of the squares as a monovariant. Before an operation it is $\sum a_i^2$. After an operation it is $\sum \left(\frac{a_i + a_j + a_k}{3}\right)^2$ where the $i, j, k$ all border a single vertex. From AM-GM or power mean this sum is at most $\sum \frac{a_i^2 + a_j^2 + a_k^2}{3} = \sum a_i^2$, with equality only when numbers across diagonals are equal. So the cube must be a b b a on the top and b a a b on the bottom or similar if it is to work. One step symmetrically switches everything, and another turns it back. So ten steps with $a \not= b$ yields an identity transformation, so the answer is no; however, we have classified all such cubes above, if we include the possibility that $a = b$ as well. I'm pretty sure the result is also true in fields of characteristic $0$ but don't cite me on this (and anyway you have to be careful about using monovariants, since you need an ordering first; I think the bash linear method, though, proves it). Actually, considering a cube as two connected tetrahedron shapes that alternate as we do the average property helps a lot in visualizing this as well as actually proving this stuff, I think. Like $\frac{S - \frac{S - x}{3}}{3} = x$ so $x = \frac{S}{4}$ where $S$ is the sum of the tetrahedral guys. Using this in fact does clear up the whole field thing, by making the equations easy and accessible.