$\def\RR{\mathbb{R}}$A brute force approach shows that there are no other $C^2$ solutions. Let $F: \RR^n \to \RR^n$ have orthogonal Jacobian everywhere. We will show that the Hessian of $F$ vanishes everywhere, so $F$ is linear.

It is enough to show that Hessian vanishes at $0$, since there is nothing special about $0$.
Translating and rotating our coordinates, we may assume that $F(0) =0$ and the Jacobian at $0$ is the identity. So, writing the components of $F$ as $(F_1, \ldots, F_n)$, we have
$$F_j(x_1, \ldots, x_n) = x_j + \sum_{a,b} Q^j_{ab} x_a x_b + (\mbox{higher order terms}).$$
Here $Q^1$, $Q^2$, ..., $Q^n$ are each symmetric $n \times n$ matrices. Our goal is to show $Q^j=0$.

Up to linear terms, the $(i,j)$ entry in the Jacobian is $\delta_i^j + 2 \sum_k Q^j_{ik} x_k$. Writing down the condition that the $j$-th column has length $1$, up to linear terms, gives $1+2 \sum Q^j_{jk} x_k = 1$. So $Q^j_{jk}=0$ and, by the symmetry of $Q^j$, we also have $Q^j_{kj}=0$.

Let $i \neq j$. Writing down the condition that the $i$-th and $j$-th column are orthogonal, up to linear order, gives $2 \sum_k Q^i_{jk} x_k + 2 \sum_k Q^j_{ik} x_k=0$, so $Q^j_{ik} = - Q^i_{jk}$ whenever $i \neq j$.

If $(i,j,k)$ are all distinct, we have $Q^i_{jk} = - Q^j_{ik} = Q^k_{ij} = - Q^i_{kj} = - Q^i_{jk}$. So $Q^i_{jk}=0$.

If $j$ and $k$ are distinct, we have $Q^j_{kk} = Q^k_{jk}=0$.

In all cases, we have shown the entries of $Q$ are $0$.