Are there combinatorial games of finite order different from $1$ or $2$?

Question

Are there any combinatorial games whose order (in the usual addition of combinatorial games) is finite but neither $1$ nor $2$?

Finding examples of games of order $2$ is easy (for example any impartial game), but I have not been able to think up an example with finite order where the order did not come from some sort of symmetry (for example even though Domineering is not impartial, it is easy to see that any square board will give a game of order $1$ or $2$), and such a symmetry only gives $1$ or $2$ as the possible orders.

Answer 1

There are games of order $4$ such as $$A=\{1|0\}+\{*|-1\}$$ since $A+A=*$ and so $A+A+A+A=0$.

Answer 2

Chapter III section 3 of Siegel's "Combinatorial Game Theory" contains a straightforward construction to make games with orders that are arbitrarily high powers of $2$.

If $G$ has finite order, and $n$ is three times the birthday of $G$, then $h(G)=\{n+G|-n\}$ satisfies $h(G)+h(G)=G$. This isn't too hard to prove if you are used to standard CGT proof techniques: assume $G$ is in canonical form and look at the game $h(G)+h(G)-G$ and a few cases. $n$ is made large enough that when you're looking at a subposition of $G+G+G$ you don't have to worry about which subposition it is.

This construction turns $0$ into $*=\{0|0\}$, $*$ into $\{3*|-3\}$ (which has order 4), which becomes $\{18|12*\|-15\}$ (order $8$)$\ldots$

What about odd orders? That's another question.