Let $U$ be a linear space over a division ring $D$, $G_1$ a subgroup of $GL(U)$, and $\Gamma$ a subgroup of a symmetric group $S_k$ on $\{1,\ldots,k\}, k>1$. The cartesian product $U^k=V_1$ can be regarded as a linear space over $D$, and we write any vector $v\in V_1$ in the form $v=(u_1,\ldots,u_k)$, $u_j\in U$. For any $f_1,\ldots,f_k\in G_1$ and $s\in \Gamma$, we define a mapping $f:V_1\rightarrow V_1$, $f=\langle f_1,\ldots,f_k,s\rangle$, by setting $$f(v)=f(u_1,\ldots,u_k)=\overline{v} \in V_1,$$ where the $s(v)$th component of $\overline{v}$ is $f_v(u_v)$, $v=1,\ldots,k.$ Obviously $f$ is an automorphism of $V_1$. The group of all such automorphisms is called the wreath product of the skew linear group $G_1$ and the permutation group $\Gamma$, and is denoted by $G_1\wr \Gamma$.
Recall that a monomial matrix is a square matrix with exactly one non-zero entry in each row and column. Assume that $F$ is a field. It is not hard to see that the set of all $n\times n$ monomial matrices over $F$ is conjugate to $F^*\wr S_n$, when $n>1$, when $S_n$ is the symmetric group over $n$ elements. We know that for $n>5$, $S_n$ is not solvable. Since $F^*\wr S_n$ has a copy of $S_n$, therefore for $n>5$, $F^*\wr S_n$ is not solvable. Is it true that $F^*\wr S_n$ is a solvable group, when $n\leq 5$?