lie algebras - Automorphism group of $sl_2$ over a finite field

Let $K$ be a finite field of characteristic $\geq 5$ and $\mathfrak{L} = \mathfrak{sl}_2(K)$ be the set of $2 \times 2 $ trace zero matrices over $K$. Let $H_0 = \Bigg\langle \begin{pmatrix}

1 & 0 \\
0 & -1 \\
\end{pmatrix}
\Bigg \rangle _ K$. Of course, $H_0$ is an abelian Cartan subalgebra of $\mathfrak{L}$. Let $G'$ be the Chevalley group, that is, the group of automorphisms of $\mathfrak{L}$ generated by all $exp(ad x_\alpha)$, $\alpha \neq 0$ is a root.

According to Seligman(Theorem III.4.1, Modular Lie algebra 1967), for any abelian Cartan subalgebra $H$ of $\mathfrak{L}$, there exists $\sigma \in G'$ such that $\sigma(H_0) = H$. In realization, let $E_{ij}$ be the matrix whose position $(i, j)$ is $1$ and zero elsewhere. Then $G'$ is the image of the group generated by the $I + \lambda E_{ij}$, $\lambda \in K$, under the mapping $U \mapsto \sigma_U$, where $\sigma_U: X \mapsto U^{-1} X U$ and we also have $G' \cong PSL(K)$.

Now, let $K = \mathbb{Z}_7 $ and $H = \Bigg\langle \begin{pmatrix}
0 & 1 \\
-1 & 0 \\

\end{pmatrix}
\Bigg \rangle _ K$. This is an abelian Cartan subalgebra of $\mathfrak{L}$. However, I can not find an invertible matrix $U$ over $K$ such that $U^{-1}H_0U = H$. I think it is because $-1$ is non square in $K$.

Where is my mistake?