This answer builds upon Sam Hopkins' suggestion of an oriented toroidal grid, but it's more than a comment. Bottom line: need the grid to be $2^n$ by $2^n$ to get the requested involution property.

Each edge belongs to a unique shortest cycle (USC) that wraps around the grid horizontally or vertically. By edge-transitivity, the horizontal USCs must have the same length as the vertical ones, so the grid is an $n$ by $n$ square. Need $n>=3$ since otherwise multiple edges between pairs of vertices. 

WLOG, using two USCs as axes, label the nodes as $(x,y)$ for $x,y\in\{0,...,n-1\}$, and the edges similarly as $((w,x),(y,z))$.

It's easy to see that the action of an AM on one edge defines its action on the whole graph, so the number of AMs is equal to the number of edges $= 2n^2$. Half the AMs are just translations, where the USCs of an edge and of its image are parallel. The others are flips, where the two USCs *do* meet at one vertex.

Define three AMs $a,b,c$ by their action on the edge $((0,0),(1,0))$ as $((1,0),(2,0))$, $((0,1),(1,1))$, $((0,0),(0,1))$ respectively. So $a$ is "right", $b$ is "up" and $c$ is "x-y flip". 

$a^n = b^n = c^2 = 1, ab = ba, cac = b, cbc = a$.

Let $d = ab = ba$ (called "diagonal"). Then $cd = dc$.

$(c{a^p}{b^q})^2 = d^{(p+q)}$. So $c{a^p}{b^q}$ has even order which divides $2n$. There are hence exactly $n$ flip involutions: $c{a^p}{b^{(n-p)}}$. Write them as $cg^p$, where $g=ab^{-1}$.

Suppose $n$ has an odd factor $k$, so $n = km$. Then any AM of the form $c{g^p}{d^m}$ taken to the $k$th power gives ${(cg^p)^k}*{d^{mk}} = cg^p$. So these involutions *are* powers of other AMs, contrary to the request.

On the other hand, suppose $n$ is a power of $2$, and that a flip involution $f = (g^{2^j})$ for some AM $g$. Then $f = (g^2)^{2(j-1)}$. But $g^2$ is a translation, not a flip, so $f$ cannot be a flip either. So to get the requested property we need that the grid be square of size a power of $2$.