This follows from Lemma \ref{lemma-branches} but we will also give a direct proof. Denote $\mathfrak m$ the maximal ideal of the ring $A$. Recall that the residue field $\kappa = A/\mathfrak m$ is the same as the residue field of $A^h$. \medskip\noindent Assume (2). Let $\mathfrak p^h$ be the unique minimal prime of $A^h$. The flatness of $A \to A^h$ implies that $\mathfrak p = A \cap \mathfrak p^h$ is the unique minimal prime of $A$ (by going down, see Algebra, Lemma \ref{algebra-lemma-flat-going-down}). Also, since $A^h/\mathfrak pA^h = (A/\mathfrak p)^h$ (see Algebra, Lemma \ref{algebra-lemma-quotient-henselization}) is reduced by Lemma \ref{lemma-henselization-reduced} we see that $\mathfrak p^h = \mathfrak pA^h$. Let $A'$ be the integral closure of $A/\mathfrak p$ in its fraction field. We have to show that $A'$ is local. Since $A \to A'$ is integral, every maximal ideal of $A'$ lies over $\mathfrak m$ (by going up for integral ring maps, see Algebra, Lemma \ref{algebra-lemma-integral-going-up}). If $A'$ is not local, then we can find distinct maximal ideals $\mathfrak m_1$, $\mathfrak m_2$. Choose elements $f_1, f_2 \in A'$ with $f_i \in \mathfrak m_i$ and $f_i \not \in \mathfrak m_{3 - i}$. We find a finite subalgebra $B = A[f_1, f_2] \subset A'$ with distinct maximal ideals $B \cap \mathfrak m_i$, $i = 1, 2$. Note that the inclusions $$ A/\mathfrak p \subset B \subset \kappa(\mathfrak p) $$ give, on tensoring with the flat ring map $A \to A^h$ the inclusions $$ A^h/\mathfrak p^h \subset B \otimes_A A^h \subset \kappa(\mathfrak p) \otimes_A A^h \subset \kappa(\mathfrak p^h) $$ the last inclusion because $\kappa(\mathfrak p) \otimes_A A^h = \kappa(\mathfrak p) \otimes_{A/\mathfrak p} A^h/\mathfrak p^h$ is a localization of the domain $A^h/\mathfrak p^h$. Note that $B \otimes_A \kappa$ has at least two maximal ideals because $B/\mathfrak mB$ has two maximal ideals. Hence, as $A^h$ is henselian we see that $B \otimes_A A^h$ is a product of $\geq 2$ local rings, see Algebra, Lemma \ref{algebra-lemma-mop-up}. But we've just seen that $B \otimes_A A^h$ is a subring of a domain and we get a contradiction. \medskip\noindent Assume (1). Let $\mathfrak p \subset A$ be the unique minimal prime and let $A'$ be the integral closure of $A/\mathfrak p$ in its fraction field. Let $A \to B$ be a local map of local rings inducing an isomorphism of residue fields which is a localization of an \'etale $A$-algebra. In particular $\mathfrak m_B$ is the unique prime containing $\mathfrak m B$. Then $B' = A' \otimes_A B$ is integral over $B$ and the assumption that $A \to A'$ is local implies that $B'$ is local (Lemma \ref{lemma-local-tensor-with-integral}). On the other hand, $A' \to B'$ is the localization of an \'etale ring map, hence $B'$ is normal, see Algebra, Lemma \ref{algebra-lemma-normal-goes-up}. Thus $B'$ is a (local) normal domain. Finally, we have $$ B/\mathfrak pB \subset B \otimes_A \kappa(\mathfrak p) = B' \otimes_{A'} (\text{fraction field of }A') \subset \text{fraction field of }B' $$ Hence $B/\mathfrak pB$ is a domain, which implies that $B$ has a unique minimal prime (since by flatness of $A \to B$ these all have to lie over $\mathfrak p$). Since $A^h$ is a filtered colimit of the local rings $B$ it follows that $A^h$ has a unique minimal prime. Namely, if $fg = 0$ in $A^h$ for some non-nilpotent elements $f, g$, then we can find a $B$ as above containing both $f$ and $g$ which leads to a contradiction.