My four prism is represented by the vertices $(\pm1, 1, 0), (0, 1, \pm1), (\pm1, -1, 0), (0, -1, \pm1)$. So it is like a rectangular prism with squares at the ends. I'm trying to show that the group of 16 matrices expressed by $$ \begin{bmatrix} \pm1 & 0 & 0 \\ 0 & \pm1 & 0 \\ 0 & 0 & \pm 1 \\ \end{bmatrix} \cup \begin {bmatrix} 0 & 0 & \pm1 \\ 0 & \pm1 & 0 \\ \pm1 & 0 & 0 \\ \end{bmatrix} $$ is a symmetry group of this four prism.
Could I do this by showing which symmetry of the prism each matrix represents? Such as the matrix \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} is a reflection through $y=0$, but do this with each of the 16? Is there a more efficient/methodological way of doing this?
Let us fix one vertex and focus on it. First note that all of the vertices are symmetric so that in principle our vertex may be sent to any other. This gives 8 possible choices for the image of our vertex under a symmetry. Furthermore, a symmetry must preserve adjacency so that the three edges coming off of our vertex need to be specified. But 2 of these have length $\sqrt{2}$ and one has length 2 and so the only choice left is permuting the two sides which have the same length. Therefore, the symmetry group has at most 16 elements (8 choices for the images of the vertex and 2 choices for the images of the edges). At this point, you can directly verify that the two matrices given are symmetries (i.e. they send the vertices to themselves and so by linearity they send the edges to themselves) and that they generate a group of order 16 so that these are in fact all of the symmetries of the given object.