By a "square rectangular cuboid", I mean this particular shape...
1: https://i.stack.imgur.com/yxcKDm.jpg
...made of two squares and four rectangles. Also known as a square rectangular parallelpiped.
Define the group $G$ to be strictly the rotational symmetry group of this above shape. As far as I'm aware, there are only three axes of symmetry (the x, y and z axes). If we take the vertical axis to be the y-axis, then (correct me if I'm wrong) there are y-axis rotations by $90^{\circ}$, $180^{\circ}$, $270^{\circ}$ and $360^{\circ}=\mathbf{1}$. About the x axis there is a $180^\circ$ rotation and $\mathbf{1}$ for the $360^\circ$ rotation, and similarly for the z-axis rotations we have the same thing, a $180^\circ$ rotation and $\mathbf{1}$ for the $360^\circ$ rotation. So this group appears to have order 6 (omitting the repeated $\mathbf{1}$ elements obviously).
I have read that this is isomorphic to the entire symmetry group of the square which includes both rotations and reflections. However, that group has order 8, so since a group isomorphism has to include a bijection, I don't see how this can be true.
Is there something I'm missing here or not understanding?
To expand on the comment—you're missing the elements of the rotational symmetry group that are the composition of a 180° rotation about the $x$ or $z$ axis and a 90° rotation about the $y$ axis.
To see the group isomorphism intuitively, you can think about identifying the vertical edges of length $c$ in this figure with the vertices of a square. By extruding the vertices of the square into edges, the reflections in the square's symmetry group become the 180° rotations about the $x$ and $z$ axes in this figure's rotational symmetry group.