TITLE: Union of conjugacy classes of $O(n)$ is not a subgroup
QUESTION [5 upvotes]: Let $O(n)$ be the standard orthogonal group of real matrices. I am trying to prove the following: 
$N = \bigcup_{g\in GL_n(\mathbb{R})}g\cdot O(n)\cdot g^{-1}$ is not a subgroup of $GL_n(\mathbb{R})$.
I know that if it was a subgroup then it was equal to the normal closure of $O(n)$ but I do not know what that is...
Motivation: 
It is proved here that a linear automorphism $T:V \rightarrow V$ preserves some inner product on $V$ if and only if the matrix of $T$ w.r.t an arbitrary basis is similar to an orthogonal matrix. I want to prove a composition of two transformation of this type is not necessarily also of that type. (Which amounts to proving $N$ is not a subgroup, since closure under taking inverses clearly holds).

REPLY [4 votes]: Note that $N$ is the set of real diagonalizable matrices that only have (complex) eigenvalues of magnitude $1$.
In the case of $n =2$: let
$$
A = \pmatrix{0&2\\-1/2&0}
$$ 
Note that both $A$ and $A^T$ are elements of $N$.  However, the matrix $AA^T$ has eigenvalues $4,1/4$, which are not of magnitude $1$.
You can generalize this counterexample by considering the block-matrix
$$
\pmatrix{A & 0\\0 & I_{n-2}}
$$
Note that this counterexample still works if you restrict to the special orthogonal group.