First assume that $k$ is algebraically closed. In this case the equivalence of (1) and (3) is trivial. The equivalence of (1) and (3) with (2) holds because the only nondegenerate quadric in two variables is $xy$ up to change in coordinates. The equivalence of (1) and (6) is  Lemma \ref{lemma-multicross-algebra}. After replacing $X$ by an affine neighbourhood of $x$, we may assume there is a closed immersion $X \to \mathbf{A}^n_k$ mapping $x$ to $0$. Let $f_1, \ldots, f_m \in k[x_1, \ldots, x_n]$ be generators for the ideal $I$ of $X$ in $\mathbf{A}^n_k$. Then $\Omega_{X/k}$ corresponds to the $R = k[x_1, \ldots, x_n]/I$-module $\Omega_{R/k}$ which has a presentation $$ R^{\oplus m} \xrightarrow{(\partial f_j/\partial x_i)} R^{\oplus n} \to \Omega_{R/k} \to 0 $$ (See Algebra, Sections \ref{algebra-section-differentials} and \ref{algebra-section-netherlander}.) The first Fitting ideal of $\Omega_{R/k}$ is thus the ideal generated by the $(n - 1) \times (n - 1)$-minors of the matrix $(\partial f_j/\partial x_i)$. Hence (2), (4), (5) are equivalent by Lemma \ref{lemma-fitting-ideal} applied to the completion of $k[x_1, \ldots, x_n] \to R$ at the maximal ideal $(x_1, \ldots, x_n)$. \medskip\noindent Now assume $k$ is an arbitrary field. In cases (2), (4), (5), (6) the residue field $\kappa(x)$ is separable over $k$. Let us show this holds as well in cases (1) and (3). Namely, let $Z \subset X$ be the closed subscheme of $X$ defined by the first Fitting ideal of $\Omega_{X/k}$. The formation of $Z$ commutes with field extension (Divisors, Lemma \ref{divisors-lemma-base-change-and-fitting-ideal-omega}). If (1) or (3) is true, then there exists a point $\overline{x}$ of $X_{\overline{k}}$ such that $\overline{x}$ is an isolated point of multiplicity $1$ of $Z_{\overline{k}}$ (as we have the equivalence of the conditions of the lemma over $\overline{k}$). In particular $Z_{\overline{x}}$ is geometrically reduced at $\overline{x}$ (because $\overline{k}$ is algebraically closed). Hence $Z$ is geometrically reduced at $x$ (Varieties, Lemma \ref{varieties-lemma-geometrically-reduced-upstairs}). In particular, $Z$ is reduced at $x$, hence $Z = \Spec(\kappa(x))$ in a neighbourhood of $x$ and $\kappa(x)$ is geometrically reduced over $k$. This means that $\kappa(x)/k$ is separable (Algebra, Lemma \ref{algebra-lemma-characterize-separable-field-extensions}). \medskip\noindent The argument of the previous paragraph shows that if (1) or (3) holds, then the first Fitting ideal of $\Omega_{X/k}$ generates $\mathfrak m_x$. Since $\mathcal{O}_{X, x} \to \mathcal{O}_{X_{\overline{k}}, \overline{x}}$ is flat and since $\mathcal{O}_{X_{\overline{k}}, \overline{x}}$ is reduced and has depth $1$, we see that (4) and (5) hold (use Algebra, Lemmas \ref{algebra-lemma-descent-reduced} and \ref{algebra-lemma-apply-grothendieck}). Conversely, (4) implies (5) by Algebra, Lemma \ref{algebra-lemma-criterion-reduced}. If (5) holds, then $Z$ is geometrically reduced at $x$ (because $\kappa(x)/k$ separable and $Z$ is $x$ in a neighbourhood). Hence $Z_{\overline{k}}$ is reduced at any point $\overline{x}$ of $X_{\overline{k}}$ lying over $x$. In other words, the first fitting ideal of $\Omega_{X_{\overline{k}}/\overline{k}}$ generates $\mathfrak m_{\overline{x}}$ in $\mathcal{O}_{X_{\overline{k}, \overline{x}}}$. Moreover, since $\mathcal{O}_{X, x} \to \mathcal{O}_{X_{\overline{k}}, \overline{x}}$ is flat we see that $\text{depth}(\mathcal{O}_{X_{\overline{k}}, \overline{x}}) = 1$ (see reference above). Hence (5) holds for $\overline{x} \in X_{\overline{k}}$ and we conclude that (3) holds (because of the equivalence over algebraically closed fields). In this way we see that (1), (3), (4), (5) are equivalent. \medskip\noindent The equivalence of (2) and (6) follows from Lemma \ref{lemma-2-branches-delta-1}. \medskip\noindent Finally, we prove the equivalence of (2) $=$ (6) with (1) $=$ (3) $=$ (4) $=$ (5). First we note that the geometric number of branches of $X$ at $x$ and the geometric number of branches of $X_{\overline{k}}$ at $\overline{x}$ are equal by Varieties, Lemma \ref{varieties-lemma-geometric-branches-and-change-of-fields}. We conclude from the information available to us at this point that in all cases this number is equal to $2$. On the other hand, in case (1) it is clear that $X$ is geometrically reduced at $x$, and hence $$ \delta\text{-invariant of }X\text{ at }x \leq \delta\text{-invariant of }X_{\overline{k}}\text{ at }\overline{x} $$ by Varieties, Lemma \ref{varieties-lemma-delta-invariant-and-change-of-fields}. Since in case (1) the right hand side is $1$, this forces the $\delta$-invariant of $X$ at $x$ to be $1$ (because if it were zero, then $\mathcal{O}_{X, x}$ would be a discrete valuation ring by Varieties, Lemma \ref{varieties-lemma-delta-invariant-is-zero} which is unibranch, a contradiction). Thus (5) holds. Conversely, if (2) $=$ (5) is true, then assumptions (a), (b), (c) of Varieties, Lemma \ref{varieties-lemma-geometrically-normal-in-codim-1} hold for $x \in X$ by Lemma \ref{lemma-2-branches-delta-1}. Thus Varieties, Lemma \ref{varieties-lemma-delta-invariant-and-change-of-fields-better} applies and shows that we have equality in the above displayed inequality. We conclude that (5) holds for $\overline{x} \in X_{\overline{k}}$ and we are back in case (1) by the equivalence of the conditions over an algebraically closed field.