From the limited amount of context you give, I am guessing that ##H## is supposed to be the Higgs field? If so, then ##H(x)## is a scalar field, so by definition, ##H(x)## is already invariant under rotations and translations. As a consequence, ##H^\dagger H## is also invariant under these.
Note that ##SU(2)\times U(1)## is a gauge or "internal" symmetry group. It therefore has nothing at all to do with rotations and translations (collectively called the Poincare group). So the transformation properties under ##SU(2)\times U(1)## are completely independent from the transformation properties under the Poincare symmetries. The latter are determined by the type of field we're dealing with: scalar, spin 1/2, vector, etc.