When $A\in M_n(\mathbb{Q})$, we consider the pencil $A-xA^T$. Then $p_A(x)=\det(A-xA^T)$ is a self-reciprocal polynomial. $p_A$ can only be irreducible if $n=2p$ is even.
Question: For every $p$, does there exist a matrix $A\in M_{2p}(\mathbb{Q})$ such that $\mathrm{Gal}(p_A)=C_2\wr S_p$ (which is the maximal Galois group for a self-reciprocal polynomial; it contains a subgroup isomorphic to $S_p$ and has cardinality $2^pp!$) ?
I am sure that the answer is yes; still have to show it !
A candidate family $(A_{2p})$ is as follows.
Let $A_{n}=[a_{i,j}]\in M_n(\mathbb{Q})$, where $a_{i,i}=i,a_{i,i+1}=1$ and otherwise, $a_{i,j}=0$.
$f_{2p}(x)=\det(A_{2p}-x{A_{2p}}^T)$, where $A_{2p}-x{A_{2p}}^T$ is a 1-band matrix.
One has the recurrence formula $f_n(x)=n(1-x)f_{n-1}(x)+xf_{n-2}(x)$.
Experiments seem to "show" that it works. What do you think ? Thanks in advance.
ANSWER to RVD de Bruyn. There are $2$ problems.
$\bullet$ Is $C_2\wr S_p$ realizable as a Galois group of a polynomial in $\mathbb{Q}[x]$ ?
$\bullet$ Can we write such a polynomial in the form $\det(A-xA^T)$ where $A$ is a rational matrix ?
For realization part, there are 2 standard methods
i) We show the required result over $\mathbb{Q}[T]$ (generic polynomial)and we use the Hilbert irred. Theorem to deduce the existence of a suitable specialization (obviously we do not obtain explicitly the polynomial, but no matter).
ii) We explicitly find a suitable family of polynomials. This is what I propose with my $(A_{2p})$'s; but I don't know how to do it.
In general, what is easier? Method i) or ii)? For $S_p$,both methods work. For i), we highlight the presence of a generator of $S_p$ (cf. wiki). For ii) we use Osada's example $x^p-x-1$.
Remark. $C_2\wr S_p$ admits a generator with $3$ elements: a cycle of order $2$, a product of 2 disjoint cycles of order 2, a product of 2 disjoint cycles of order $p$.
I think that $C_2\wr S_p$ is in the list of realizable groups, but, unfortunately, this is not enough to deduce the requested result
