Combinatorial structure of the tesseract

Question

On a cube, you can define the notion of "the next edge around this vertex, in clockwise order."

Formally, if $D$ is the space of darts (edges of the cube with one of the endpoint vertices marked as special), this "next edge" function is a permutation $\sigma:D\rightarrow D$ that keeps the special vertex the same. If you also let $\alpha:D\rightarrow D$ be the involution that flips which endpoint is considered special, you get a map $\alpha\circ \sigma$ whose orbits make a nice picture:

Now I am trying to do the same thing on a tesseract, but I don't know how to define the "next edge around this vertex" function $\sigma:D\rightarrow D$. I believe the tesseract is orientable, so it should be possible to choose a $\sigma$ with the right combinatorial properties (even though there's no cross product in 4D). According to my math*, the resulting map $\alpha\circ \sigma$ should have eight orbits (compared to the cube's four orbits shown above).

My question is this: Given an edge of the tesseract with one endpoint marked as special, how do I define "the next edge" with that same endpoint?

More precisely, I think my question is:

Question: If $D$ is the space of darts (edges of the tesseract with one of the endpoints marked as special), and $\alpha:D\rightarrow D$ is the involution that changes which endpoint is special, is there a permutation $\sigma:D\rightarrow D$ such that:

1. The cycles of $\sigma$ are exactly the sets of darts with a common special endpoint. Intuitively, $\sigma$ acts like it "pivots" darts around their special endpoint.
2. The orbits of $(\alpha \circ \sigma^{-1} \circ \alpha \circ \sigma)$ all have the same size, and there are 2*8 = 16 orbits. (I suppose this means that since there are 32 edges = 64 darts, each orbit is a cycle of 4 points.)

Those are the two main constraints. A bonus constraint, sort of difficult to articulate formally, is that I would expect $\sigma$ to interact nicely with the symmetries/rotations of the tesseract. (I think I mean that for every rotation $\rho$ of the tesseract, $\rho^{-1}\sigma\rho = \sigma$.)

I am hoping for a concrete answer, perhaps even in terms of a specific tesseract such as $\langle \pm 1, \pm 1, \pm 1, \pm 1\rangle$. I would like to understand how unique the answer is---but it is ok if the answer is not unique as long as it has the required combinatorial structure.

* Note on my math: If $T$ is the Tutte polynomial of the tesseract graph, then $1+\log_2 |T(-1,-1)|$ is the number of strands in the knotwork, i.e. the desired number of orbits of this permutation $\alpha\circ\sigma$.