= Solution
Denote the four listed matrices, in order, by $I,P,Q,W$. As <permutation matrices>, $P,Q,W$ act on the four basis vectors by $(12)(34)$, $(13)(24)$ and $(14)(23)$, respectively. Direct multiplication gives
$$
P^2=Q^2=W^2=I,\qquad PQ=QP=W,\quad QW=WQ=P,\quad WP=PW=Q.
$$
These relations prove closure and show that each element is its own inverse. The identity is included and matrix multiplication is associative, so the matrices form a group.
The relations coincide with those of the <Klein four-group> $D_2$: take its half-turn generator to $P$, a reflection generator to $Q$, and their product to $W$. All four matrices are distinct, so this assignment is a <faithful representation> of $D_2$. It is the <four-point regular representation of the Klein four-group>.
Every matrix fixes $v=(1,1,1,1)^T$, since it permutes its coordinates. Therefore
$$
\boxed{\operatorname{span}\{(1,1,1,1)^T\}}
$$
is a one-dimensional <invariant subspace>, nonzero and strictly smaller than the four-dimensional space. This proves \b[the representation is reducible]. In fact the four mutually orthogonal sign vectors $(1,1,1,1)$, $(1,1,-1,-1)$, $(1,-1,1,-1)$ and $(1,-1,-1,1)$ give invariant lines and completely diagonalize the representation, as in the normalized <Hadamard matrix> construction.
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