If I have a natural number $o=2n$ or $4n$ I can create a polyhedron whose group of symmetries has order $o$ by making a polygon like $C_n$ and then dragging it out to make a prism (I believe this is dihedral symmetry in the case $o=4n$; you need colors in the $o=2n$ case, not sure what this is called).
What if $o=3n$? Is the group $C_n \times C_3$ isomorphic to the group of symmetries for any polyhedron? It seems like there should be some clever way to make a pyramid-type structure, but I can't quite figure it out.
If you're looking for a way to get the factor of $3$ in a similar systematic way as you got the factors of $2$ and $4$, there isn't one. The $2$ and $4$ are related to reflections, which are of order $2$. You can read more here and here about why there are sequences of point groups of orders $2n$ and $4n$ but not $3n$.
However, if you're just looking for any polyhedron whose symmetry group is of order $3n$, or in fact of any order $k$ you like, you can use the cyclic groups $C_k$, which exist for all orders. Take a regular $k$-gon and break the reflection symmetries without breaking the rotational symmetry, for instance by cutting off all the corners in the same asymmetric way. Then form a pyramid over the resulting polygon with a point above the centre. The resulting polyhedron has rotational symmetry of order $k$ and nothing else.