For a polynomial equation in one variable over $\mathbb{Q}$, it is well known that the equation is solvable by radicals if and only if the equation's Galois group (which is a finite group) is solvable. The 'only if' part is important - we need it to prove that an equation of the fifth degree exists which is not solvable by radicals.
For differential equations, we have the following theorem, which I quote from [1] (theorem 2.64, page 151):
If an n-th order ordinary differential equation admits an n-parameter solvable group of symmetries, then the equation is solvable by quadratures - that is, the general solution to the equation can be expressed by integrals.
Here by an 'n-parameter solvable group' we mean a Lie group of dimension n which is solvable. By the equation "admitting" the group we mean that the group, acting on $\mathbb{R}^2=X \times Y$ (where $x$ is the independent variable in the equation and $y$ is the dependent variable), transforms solutions of the equation to other solutions of the equation.
Is the converse to the above theorem true? that is, If an n-th order ordinary differential equation is not solvable by quadratures, does it follow that the equation does not admit a solvable n-parameter group of symmetries?
If this converse is not true, is it because there are differential equations solvable by quadratures which do not admit an n-parameter group of symmetries at all? Or is it because there are differential equations solvable by quadratures which admit a nonsolvable n-parameter group of symmetries? Whatever of these cases is correct, is there a concrete example of such a differential equation?
[1] Olver, Peter J., Applications of Lie groups to differential equations., Graduate Texts in Mathematics. 107. New York: Springer-Verlag. xxviii, 513 p. (1993). ZBL0785.58003.
