a symmetry is always a transformation that leaves a given object unchanged.
To reveal an analogy, consider a dodecahedron centered on the origin in flat Euclidean space. An isometry that permutes the dodecahedron's vertices with each other is a symmetry of the dodecahedron, according to the usual meaning of "symmetry." The symmetry transformation leaves the whole dodecahedron unchanged, even if it does not leave the individual vertices unchanged.
I'm curious why observable symmetries of physical systems are exactly those transformations that map solutions of the equation of motion into other solutions.
In physics we are often more concerned with a system's behavior than with an individual static object. The equations of motion are a concise way of describing which behaviors are allowed. A symmetry transformation leaves the whole set of solutions unchanged, even if it does not leave the individual solutions unchanged. A symmetry transformation may permute different solutions with each other, but it transforms each valid solution into another (possibly different) valid solution -- just like a symmetry of the dodecahedron transforms one vertex of the dodecahedron to another (possibly different) vertex of the dodecahedron.
Is there possibly a little thought experiments that motivates this definition?
The motivation was described above. To illustrate it, consider a simple example: Maxwell's equations in flat spacetime, in a classical world with no charges or currents. Maxwell's equations delineate the set of allowed behaviors of the electromagnetic field. Any behavior of the electromagnetic field, whether allowed or not, is described by giving the Faraday tensor $F_{ab}(x)$ as a smooth function of the spacetime coordinates $x$, with $F_{ab}(x)=-F_{ba}(x)$. (This antisymmetry condition is normally considered to be part of the definition of the Faraday tensor, not one of the equations of motion. More precisely, the quantities $F_{ab}$ are the components of a two-form.) A given behavior of the field $F_{ab}(x)$ is "physically allowed" (in this toy model) if and only if it satisfies Maxwell's equations
$$
\partial^a F_{ab}=0
\hskip2cm
\partial_a F_{bc}+
\partial_b F_{ca}+
\partial_c F_{ab} =0.
\tag{1}
$$
One way of transforming the field is to consider any diffeomorphism (smooth map) of the spacetime manifold to itself and its corresponding pullback to the set of tensors. The quantities $F_{ab}$ are the components of a tensor (specifically a two-form), so the pullback defines a permutation of the set of behaviors of the electromagnetic field. Most of these transformations don't preserve the set of allowed behaviors; instead, most of them convert solutions of the equations of motion (1) to non-solutions. Lorentz transformations are examples of transforms that do preserve the set of allowed behaviors, so we say that Lorentz transformations are symmetries of the equations of motion (1).
The OP used the expression "observable symmetries," and that introduces a host of interesting subtleties, because then we must distinguish between transformations whose effects are observable and transformations whose effects are not. For example, assuming as usual that spacetime manifold is $\mathbb{R}^4$, the second of equations (1) implies that any allowed $F_{ab}(x)$ can be written in terms of a gauge field $A_a(x)$ like this:
$$
F_{ab}=\partial_a A_b - \partial_b A_a.
\tag{2}
$$
(This is Poincaré's lemma.)
The gauge field provides a redundant description of the Faraday tensor, in the sense that any transformation of the form
$$
A_a\to A_a+\partial_a\theta,
\tag{3}
$$
for any smooth function $\theta(x)$, leaves the Faraday tensor $F_{ab}$ invariant. The transformation (3) is a gauge transformation, and it obviously qualifies as a symmetry of the equations of motion (1), because it doesn't change the Faraday tensor $F_{ab}(x)$ at all. Physically, a gauge transformation is a trivial symmetry, because all of the model's observables can be expressed in terms of the Faraday tensor, so the gauge transformation doesn't change anything observable. (The assertion that all observables can be expressed in terms of $F_{ab}$ is part of this model's definition.)
Does a Lorentz transformation of (1) change anything observable? It generally does change the Faraday tensor $F_{ab}(x)$, of course, but with respect to what? A mathematical transform has observable effects only if it changes some relationship between physical entities; but in this toy model, the electromagnetic field is the only physical entity, isn't it? I won't try address that issue here, because I'm not sure it's within the intended scope of the OP and because this answer is already long, but I mentioned it as another example of how including the word "observable" in the question introduces some interesting subtleties.
