An operator $A$ on an $n$-dimensional real vector space is conformal in case
$$
A^T A = \alpha I \qquad \text{for some} \qquad \alpha \ge 0 ,
$$
and
$$
\mathrm{det} A \ge 0 .
$$
Let $A$ and $B$ be two conformal operators of a vector space.
Under what conditions are $A + B$ and $A - B$ again conformal?
(That is, when do the operators form an additive group?)
Some examples follow.
If $A^T = \alpha A^{-1}$ and $B^T = \beta B^{-1}$, $$ (A + B)(A^T + B^T) = (\alpha + \beta)I + AB^T + BA^T, $$ so the sum is a positive multiple of the identity just when $AB^T + BA^T = \gamma I$ for some real $\gamma$ with $\alpha + \beta + \gamma > 0$.
In the plane, this condition amounts to $a\overline b + b \overline a $ being real for two nonzero complex values $a$ and $b$, which is always the case.
But in odd dimensional spaces, $\mathrm{det}(-A) = (-1)^n \mathrm{det}A = -\mathrm{det}A$, so the conditions for conformality never hold.
With some difficulty, I have verified that, for $n = 4$, conformality is not preserved under addition. (This involved a rather complicated calculation based on a matrix representation of conformal operators.
Altogether, conformality is preserved for no odd-dimensional space, and not for $4$-dimensional space, but it is preserved for $2$-dimensional space.
What about the other even-dimensional spaces?