Identify the group of all motions that leave a right prism invariant.
It seems like the normal dihedral group, $D_n$, with respect to the base would be a subgroup of this group. That is to say that any rotation or reflection that leaves the $n$-sided polygon that makes up the base invariant in it's own right should leave the prism invariant as well. I am having some troubles figuring out what motions will leave the prism invariant that would not be a motion in $D_n$.
Is there anyone who could assist with this?
It's a little complicated. If we let $P$ be our solid prism (centered at the origin, for ease) oriented so that our $n$-gon bases are horizontal, we do have an injection $D_n \hookrightarrow \text{Sym}(P)$, and each $g \in D_n$ corresponds to a rotation about the $z$ axis, or a reflection across a vertical plane. But!
There's an additional symmetry that doesn't come from $D_n$, and opens up a whole bunch of new symmetries. It's a symmetry that interchanges the top and bottom faces, and leaves fixed (not just invariant) a plane parallel to the bases. Let's call this symmetry $\tau$, and we can even see that $(x, y, z) \overset{\tau}{\to} (x, y, -z)$.
Is it the case that, for any $g \in D_n$, we have $\tau g \in \text{Sym}(P)$ and for distinct $g_1, g_2 \in D_n$, that $\tau g_1 \neq \tau g_2$?
We also have some extraordinary symmetries when $n = 4$ and our prism is the cube. Then, $|\text{Sym}(P)|$ is $6$ times as large as $|D_n|$, which is abnormal!