I have the following question:
Does differential invariants have the same Lie symmetries?
I want to know about the relation of differential invariants of an action and their symmetry properties. Does differential invariants of a group action have isomorphic Lie symmetry group?
In [1], the authors considered the differential invariants of a fifth-order KdV types of equation (Kawaha KdV equation). Here is the Lie symmetries of the equation:
$$ X_1=\frac{\partial}{\partial x},\qquad X_2=\frac{\partial}{\partial t},\qquad X_3=at\frac{\partial}{\partial x}+\frac{\partial}{\partial u}$$
The third-order prolongation of the operator $X$ has the form
$$X^{(3)}=(c_3 at + c_1)\frac{\partial}{\partial x} + c_2 \frac{\partial}{\partial t}+c_3\frac{\partial}{\partial u}-c_3 au_x\frac{\partial}{\partial u_t}-c_3 a u_{xx}\frac{\partial}{\partial u_{xt}}-2c_3 a u_{xt} \frac{\partial}{\partial u_{tt}}-c_3 a u_{xxx}\frac{\partial}{\partial u_{xxt}}-2c_3 a u_{xxt}\frac{\partial}{\partial u_{xtt}}-3c_3 a u_{xtt} \frac{\partial}{\partial u_{ttt}}$$
In the paper a set of generating differential invariants is found. For instance $u_{xxx}$ is a differential invariant of Kawaha KdV.
Making simple calculations, one obtains
$$X^{(3)}(u_{xxx}) = 0.$$
Therefore $u_{xxx}$ is a differential invariant. However, $u_{xxx}$ has the Lie symmetries: $$ X_1=F_1(t)\partial_x,\qquad X_2=F_2(t)\partial_u,\qquad X_3=F_3(t)\partial_t,\qquad X_4=F_4(t)x^2\partial_u,\qquad X_5=F_5(t)x\partial_x,\qquad X_6=F_6(t)x\partial_u,\qquad X_7=F_7(t)u\partial_u,\qquad X_8=F_8(t)\left(\frac{1}{2}x^2\partial_x+xu\partial_u\right). $$ The Lie symmetries of $u_{xxx}$ is different from the Kawaha KdV.
In another paper [2], the concept of hidden symmetries is raised. According to the paper, a hidden symmetry is a Lie point symmetry which appears in the target differential equation after a change of order using a nonlocal transformation and which does not have a point counterpart in the source equation.
