For two linear maps $A, B:\Bbb R^n\to \Bbb R^n$, define the relation:
$$ A\sim B \qquad \iff \qquad A(D_n) = B(D_n) \tag1 $$ where $D_n$ denotes the $n$-dimensional ball: $$ D_n=\{x\in\Bbb R^n ~/~ \|x\|_2=1\} \tag2 $$
Obviously, $\sim$ is an equivalence relation, and the question arises what's a moduli space for $M/\!\sim$, $~M=\{\varphi ~/~ \varphi:\Bbb R^n\to\Bbb R^n, \varphi\text{ linear} \}$.
Now for any orthogonal $Q\in O(n,\Bbb R)$, the unit ball is invariant under the action of such $Q$, and hence
$$ Q\text{ orthogonal} ~\implies~ A\sim AQ \tag3 $$ and also $$ Q\text{ orthogonal} ~\implies~ Q\sim \operatorname{Id}_n \tag4 $$ where I think the latter could be an equivalence actually.
For the case of regular $A$, i.e. $A\in\operatorname{GL}(n,\Bbb R)\subset M$, due to $(3)$ we get
$$\begin{align} \dim(\operatorname{GL}(n,\Bbb R)/\!\sim) &= \dim(\operatorname{GL}(n,\Bbb R)) - \dim (O(n,\Bbb R)) \\ &= n^2-1 - n(n-1)/2 \\ &= n(n+1)/2 -1\\ \tag5 \end{align}$$
but for non-regular mappings, there will be lower dimensional "appendages" (borders?).