The group is isomorphic to $S_3 \times S_2$, where $S_3$ acts on $\{2,3,4\}$ and $S_2$ acts on $\{1,5\}$. In fact, symmetries with respect to planes perpendicular to the planes containing the triangles and passing by the height of the triangles swap exactly two rectangular facets, corresponding to the permutations $(23), (34), (24)$, that generate the whole $S_3$. Similarly, the symmetry with respect to the plane parallel to the triangles and passing by the centre of the solid, swaps $1$ and $5$ while fixing all the rest.

1) The conjugacy classes are products of the conjugacy classes of $S_3$ (the identity, the transpositions, and the 3-cycles) and conjugacy classes of $S_2$ (the identity and the transposition), so there are six in total: $A = \{ (234)(15), (243)(15) \}$, $B = \{ (234), (243) \}$, $C = \{ (15) \}$, $D = \{ (23)(15), (34)(15), (24)(15) \}$, $E = \{ (23), (34), (24) \}$, and $I = \{ id\}$.

In fact, if $\sigma_1$ and $\sigma_2$ are conjugate in $S_3$, and $\tau_1$ and $\tau_2$ are conjugate in $S_2$, then $(\sigma_1, \tau_1)$ is conjugate to $(\sigma_2, \tau_2)$ in $S_3 \times S_2$ (it is enough to conjugate by the same elements: if $\alpha \sigma_1 \alpha^{-1} = \sigma_2$ and $\beta \tau_1 \beta^{-1} = \tau_2$ then $(\alpha, \beta)(\sigma_1, \tau_1)(\alpha, \beta)^{-1} = (\sigma_2, \tau_2)$) and vice versa (from a conjugacy relation in the product you can obtain one in both $S_3$ and $S_2$ by projecting to the components).

2) $S(P)$ does not have any normal subgroup of size $4$. For a subgroup to be normal, it has to be union of conjugacy classes, and must contain the identity, so it can be either $A \cup C \cup I$, $B \cup C \cup I$, $D \cup I$, or $E \cup I$. None of these is a subgroup: $(234)(15) \ast (15) = (234) \not \in A \cup C \cup I$, $(234) \ast (15) = (234)(15) \not \in B \cup C \cup I$, $(24)(15) \ast (23)(15) = (234) \not \in D \cup I$, and finally $(24) \ast (23) = (234) \not \in E \cup I$.

Alternatively, since no element has order $4$ then a subgroup of order $4$ must be generated by two commuting elements of order $2$, and it still has to be union of conjugacy classes. By looking at the sizes, it must be either $D \cup I$ or $E \cup I$, but the elements in either D or E don't all commute with each other, and so they must generate a larger subgroup.