Does a crystalline ferromagnetic solid break the rotational symmetry twice?

It seems like you are confusing spatial rotation symmetry and spin rotation symmetry. They are different symmetries (and both are present in nature).

In terms of group theory we will say that spin rotation symmetry is manifested by invariance of the hamiltonian with respect to group $SU(N)_{\text{spin}}$ where $N=2S+1$. Spatial symmetry by $SO(3)_{\text{space}}$. If you have both independently then the hamiltonian is invariant with respect to $SU(N)_{\text{spin}} \times SO(3)_{\text{space}}$, any combination of the two. You can also have the hamiltonian be invariant to $SO(3)_{\text{spin,space}}$. This is a subgroup of $SU(N)_{\text{spin}} \times SO(3)_{\text{space}}$, in which rotations are taken simultaneously in space and spin space. Now most systems are symmetric with respect to $SU(N)_{\text{spin}} \times SO(3)_{\text{space}}$, however the action of physically doing a rotation corresponds to $SO(3)_{\text{spin,space}}$. This is one reason to be confused. Pure spin rotation ($SU(N)_{\text{spin}}$) can be obtained with magnetic fields.

Now we will say that a system breaks a symmetry spontaneously if the observed ground state is not symmetric with respect to a given group although the hamiltonian is. Ferromagnetic systems spontaneously break $SU(N)_{\text{spin}}$ symmetry and crystals $SO(3)_{\text{space}}$ symmetry. So, to answer your question, they do not break the same symmetry. A ferromagnetic crystal will break both.

They also break $SO(3)_{\text{spin,space}}$ although it is not necessarily the case that when the first two are broken, this last will be broken. You could imagine a system in which the spin rotation symmetry breaking perfectly compensates the spatial rotation symmetry breaking.