Let $\mathbb K$ be a field, $n\geq 1$, and $G=GL_n({\mathbb K})$ be the group of invertible $n \times n$ matrices with coefficients in $\mathbb K$. For $J\in G$, we can define (in analogy to orthogonal&symplectic matrix groups)
$$
H=\lbrace M \in G | {}^tMJM=J\rbrace
$$
This is a subgroup of $G$, and the determinant defines a homomorphism $H \to \lbrace -1,1\rbrace$ (because $({\sf det} M)^2=1$ for any $M\in H$, by taking determinants in the equation above). So the image of this homomorphism is either $\lbrace 1 \rbrace$ (trivial) or $\lbrace -1,1\rbrace$ (full).
In some cases (say $\mathbb K=\mathbb R$ and $J=I_n$ : then $H$ is the group of orthogonal matrices), the image is full (for example, ${\sf diag}(1,-1)$ has determinant $-1$).
In others (such as when $J$ is as in this recent Math.stackexchange question), the image is trivial.
The (obvious) question is : for which $J$ is the image trivial or full ?
Best Answer
I'll assume $K$ is not of characteristic $2$.
If $M$ satisfies ${}^t\!MJM=J$ then, applying transposition to the equation, one also has ${}^t\!M\,{}^t\!JM={}^t\!J$. Therefore if we denote by $H_J$ the group defined by $J$, one easily checks that $H_J=H_{J_S}\cap H_{J_A}$ where $J_S=J+{}^t\!J$ is a symmetrization of $J$, and $J_A=J-{}^t\!J$ an anti-symmetrization. The answer can therefore be deduced from that for the cases where $J$ is symmetric (the orthogonal case) and where $J$ is anti-symmetric (the symplectic case). However $J$ is no longer necessarily invertible here, so one needs to consider the radical of the bilinear form defined by $J$ as well (if the radicals of $J_S$ and $J_A$ have a non-trivial intersection, it is not even ensured that $\det M=\pm1$, but this should not happen if the original $J$ is invertible).
[Added] One possible analysis is as follows. Whenever one can decompose the space into a direct sum that is othogonal for both the bilinear forms defined by $J_A$ and $J_S$, then one can define endomorphisms independently on the two factors. In paticular, if the intersections of the radicals of $J_A$ and $J_S$ is non-trivial, it will be orthogonal for both forms to any complementary subspace, so we can do anything we like on this intersection and the identity on the complementary factor, and thus see that any nonzero determinant can be achieved in this case. On the other hand, if the radical of $J_A$ is trivial, then we have a subgroup of the symplectic group defined by $J_A$, and only the determinant $1$ can be obtained. For the remaining case let $R$ be the radical of $J_A$ (so we have $\dim R>0$), and $C$ its orthogonal complement for the orthogonal form defined by $J_S$. It may be that $R$ and $C$ have a non-trivial intersection, which is the radical of the restriction of the orthogonal form to $R$; in that case, let $R'\subseteq R$ be a complementary subspace in $R$ of $R\cap C$, and $C'\supseteq C$ the orthogonal complement for $J_S$ of $R'$. Then one can check that the whole space is the direct sum of $R'$ and $C'$, orthogonal for both forms. Now if $\dim R'>0$, in other words if $C\not\subseteq R$, then one can choose an orthogonal transformation of $R'$ with determinant $-1$ and the identity on $C'$, showing that $-1$ occurs as determinant. In the remaining case $R$ is contained in $C$, both are stable subspaces for any $M$ under consideration, and the orthogonal form induces a duality between $R$ and $V/C$ so the determinant of $M$ is the one of the transformation of $C/R$ it induces. If the symplectic form induces a non-degenerate symplectic form in this subquotient (for instance when $R=C$) then $\det M=1$ always. If not we have a space $C/R$ that carries induced orthogonal and symplectic forms, the latter having a non-trivial radical; we are back in the initial situation but with a smaller dimension, so we can continue as above until reaching a decision as to whether $-1$ occurs as determinant or not.