[OscarCunningham]

Consider the 3×3 block of live cells:

    .....
    .ooo.
    .ooo.
    .ooo.
    .....

It has no parent with the same symmetry, since every neighbour of the centre cell would be the same but a cell can't live or be born with 0 or 4 neighbours.

But it does have a parent with less symmetry:

    .....
    .....
    ooooo
    .....
    .....

[remcob]

True, and this is a counterexample to my claim that the parent would retain the symmetry. I think it can be refined though:

The parent still has hor/ver mirror symmetries, and the solution rotated is also a solution. So there is a set of (in this case two) parents that is invariant under the same symmetries.

In general, a parent should still be a parent when any of the targets symmetries are applied. (But it wouldn't necessarily be the same parent).

This can be captured in a SAT by insisting that any symmetry of a solution is also a solution, which adds a lot of new constraints and no new variables, so should help with finding.