= Solution
Suppose $f$ is sufficiently smooth and <equivariant>, $f(0,\mu)=0$, and the <group representation> on the critical space is <absolutely irreducible>. Its <linearization> is then $D_xf(0,\mu)=a(\mu)I$. Assume $a(0)=0$ and $a'(0)\ne0$. The <equivariant branching lemma> states that \b[each isotropy <subgroup> $H$ with $\dim\operatorname{Fix}(H)=1$ has a local nontrivial branch of <equilibria> with that isotropy], apart from the common origin. Conjugate isotropy <subgroups> give symmetry-related branches, so branch types are counted up to conjugacy.
To see the scalar mechanism, write the <fixed-point subspace of a group action> as $\mathbb Re$ and restrict $f(se,\mu)=e\,h(s,\mu)$. <Equivariance> makes the <fixed-point subspace of a group action> invariant. Since $h(0,\mu)=0$, write $h=s\,k(s,\mu)$; $k(0,0)=0$ and $k_\mu(0,0)=a'(0)\ne0$. The <implicit function theorem> solves $k(s,\mu(s))=0$ near zero. For small nonzero $s$, the <stabilizer subgroup> is exactly that of $e$. A <normalizer> element reversing $e$ makes $h$ odd and $\mu(s)$ even; the generic branch is then a <pitchfork bifurcation>. The lemma ensures existence, not <dynamical stability>, and does not exclude further branches of smaller isotropy.
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