Sets of matrices which are irreducible but not strongly irreducible

The answer to your second question is no.

Let $E_1$, $E_2$, $E_3$ be pairwise transverse $2$-dimensional subspaces of $\mathbb{R}^4$. Consider the following semigroup: $$ \Sigma := \{M \in \mathrm{Mat}(4,4) ; \; M(E_i)=E_i, i=1,2,3\}. $$

[Edit] The following remark will be useful: Any linear transformation $T_1 : E_1 \to E_1$ can be uniquely extended to an element of $\Sigma$. Indeed, with respect to the splitting $E_1\oplus E_2$, the space $E_3$ is the graph of some isomorphism $L \colon E_1 \to E_2$. Then $\Sigma$ is exactly the set of linear transformations $T\colon \mathbb{R}^4 \to \mathbb{R}^4$ whose block form with respect to the splitting $E_1\oplus E_2$ is: $$ T = \begin{pmatrix} T_1 & 0 \\ 0 & L T_1 L^{-1} \end{pmatrix}. $$

Now, for definiteness, we choose the following three subspaces, given in coordinates $(x_1,x_2,x_3,x_4)$ by: \begin{align*} E_1 &:= \{ x_1 = 0 , \ x_2 = x_4 \}, \\ E_2 &:= \{ x_2 = 0 , \ x_3 = x_4 \}, \\ E_3 &:= \{ x_3 = 0 , \ x_1 = x_4 \}, \end{align*}

Consider the following orthogonal matrix which permutes the $E_i$'s: $$ R := \begin{pmatrix} 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} $$ Note that $R$ preserves the orthogonal splitting $F \oplus G$, where \begin{align*} F &:= \{x_1+x_2+x_3=0, \ x_4=0\},\\ G &:= \{x_1=x_2=x_3\}, \end{align*} being a rotation of $120$ degrees on $F$, and the identity on $G$.

Let $\Gamma := \Sigma \cup \{R\}$. Then:

  1. $\Gamma$ is not strongly irreducible, since $E_1 \cup E_2 \cup E_3$ is preserved.
  2. $\Gamma$ is irreducible. Indeed:

(a) $\Sigma$ preserves no $1$- nor $3$-dimensional subspace. This is easy to check using the "useful remark" above.

(b) the only $2$-dimensional subspaces preserved by $R$ are $F$ and $G$, but these are not preserved by $\Sigma$.

  1. The semigroup generated by $\Gamma$ is infinite (even after identifying rescaled matrices).
  2. $\Gamma$ preserves no union of subspaces forming a splitting of $\mathbb{R}^4$.

Therefore $\Gamma$ is a counterexample to your proposed classification.