Yes, thanks. I adjusted the case $d=1$ accordingly.

Yes, this is what I mean. Thanks, I will also edit the question accordingly.

@Angelo Yes, I assume $X$ to be irreducible. Also, I am now assuming that the fibers are irreducible which is in your example not the case (fiber over $y$ is two points). Finally, I think that in your example $X$ is in fact complete as the union of complete varieties.

Yes, thanks. I actually want the fibers to be irreducible. Do you have a counterexample when the fibers are not equidimensional?

Thanks @VítTuček. Regarding the second point: For example the 2x2 matrix $\binom{01}{10}$ is $\mathfrak{S}_2$ invariant but not $\textrm{O}(2)$ invariant. So (2) is not automatically true.

Thanks for that wonderful answer! A last question regarding the Casimir element: I understand that the irreducible components of a representation will be eigenspaces of the Casimir operator. But is it really clear that non-isomorphic irreducibles will correspond to different eigenvalues?

I have added a counterexample where $\Phi$ is not linear.