Subgroups of $GL_n(K)$ preserving polynomials Vinberg in his algebra book, when introducing groups, mentions the following (translated from Russian):

[...] Analogously, nonsingular linear transformations of the space $K^n$ [where K is some field] that preserve a given polynomial in $n$ variables constitute a subgroup of the group $GL_n(K)$. Nonsingular linear transformations of the space $\Bbb R^n$ that preserve the polynomial $x_1^2+x_2^2+\dots +x_n^2$ are called orthogonal transformations; they form a subgroup of $GL_n(\Bbb R)$ called the orthogonal group and are denoted $O_n$.

Now I realise that $O_n$ is a group and I know that it is the group of isometries fixing the origin, but that is because I know a priori about it and that the polynomial in question is actually the Euclidean distance and therefore any transformation preserving it would be an isometry. For a general polynomial $p$ in $K[x_1,\dots, x_n]$, how is one supposed to tell what the subgroup preserving $p$ would look like, for example for $p=x_1+\dots+x_n$? Or, if symmetry of the polynomial simplifies the group somehow, for an even more general, non-symmetric one like, say, $p=\sum x_i^i$? In fact, I am still not quite sure whether it even is a subgroup (I can see how the identity and composition might belong to it, but does the inverse) or what is meant by "preserving the polynomial". Is the implication that a transformation $T$ preserves a polynomial $p$ if and only if $p(T[x_1,\dots,x_n])=p(x_1,\dots,x_n)$? And finally, does there exist a polynomial for each subgroup $H$ of $GL_n(K)$, that $H$ preserves, or are there subgroups that can't be described this way?
 A: An answer to the quickly answerable questions:
First, a proof that the non-singular $p$-preserving transformations form a subgroup. We have a polynomial $p:K^n \to K$, and $A$ is $p$-preserving if $p(Ax) = p(x)$ for all $x$. For closure under multiplication, it suffices to note that for $A,B$ in the subgroup
$$
p(ABx) = p(A(Bx)) = p(Bx) = p(x).
$$
If $A$ is an element of the group, then we have
$$
p(A^{-1}x) = p(A(A^{-1}x)) = p(x).
$$
I suspect that there exist subgroups of $GL_n(K)$ that cannot be described as the preservers of polynomials.  For instance, the unitary matrices in $GL_n(\Bbb C)$ are the preservers of the function $p(x) = |x_1|^2 + \cdots + |x_n|^2$, which is not a complex polynomial on $x_1,\dots,x_n$. Proving that there is no complex polynomial that works is a bit trickier; I suspect that we can do this, however, using the fact that the unitary matrices form a compact subset of $GL_n(\Bbb C)$.
Arguably, this is a bit of a cop-out since the $p$ described above is a real polynomial on the complex and real parts.

An attempted proof of my conjecture: we note that $A$ is a $p$-preserver iff $p(Ax) - p(x)$ is the zero-polynomial over $x_1,\dots,x_n$.  Consider the element of $(\Bbb C[a_{ij}])[x]$. That is, write each coefficient of a product of some $x_i$ as a polynomial of the entries of $A$. Setting each of these of these polynomials equal to zero gives us a system of equations
$
p_1(A) = 0 , \dots , p_m(A) = 0
$
for which $A$ is a $p$ preserver iff it is non-singular and solves the system.
In other words, the $p$-preservers (if we include the singular matrices) form a Zariski-closed subset of $GL_n(\Bbb C)$.  However, because of [INSERT ALGEBRAIC GEOMETRY FACT HERE], this implies that the $A$-preservers must be either finite or form an unbounded subset of $GL_n(\Bbb C)$.
It follows that the unitary matrices cannot be a set of polynomial preservers.
