The Lie algebra ${\mathfrak n }_{2,3}$
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Let $t$ denote an automorphism of ${\mathfrak n }_{2,3}$ and let $e_1,e_2,e_3,e_4,e_5$ be the basis given in (\[b23\]). Denote as $(t_{ij})$ the matrix of $t$ in that basis. Since $t$ leaves invariant the commutator and the center and $C^1({\mathfrak n }_{2,3})=span\,\{e_3,e_4,e_5\}$ and ${\mathfrak z }({\mathfrak n }_{2,3})=span\,\{e_4,e_5\}$, we have that $t_{ij}=0$ if $i=1,2$, $j=3,4,5$.

Notice that the Lie algebra ${\mathfrak n }_{2,3}/\zz({\mathfrak n }_{2,3})$ is isomorphic to the Heisenberg Lie algebra $\hh_1$, hence $t$ induces a automorphism $\bar{t}\in \Aut(\hh_1)$. Let $\bar{x}$ denote the image of an arbitrary element $x\in {\mathfrak n }_{2,3}$ by the canonical epimorphism ${\mathfrak n }_{2,3} \to {\mathfrak n }_{2,3}/\zz({\mathfrak n }_{2,3})$, thus $\bar{t}(\bar{x})=\overline{tx}$. From the computation $[\bar{t} \bar{e}_1, \bar{t}\bar{e}_2]=\overline{te_3}$ one gets $$\label{det}
t_{11} t_{22}- t_{12} t_{21}= t_{33}.$$

We introduce the notation for the submatrices $$A:=\left(\begin{array}{cc}t_{11}&t_{12}\\ t_{21}& t_{22}\end{array}\right)\qquad B:=\left(\begin{array}{cc}t_{44}&t_{45}\\ t_{54}& t_{55}\end{array}\right)
.$$ Note that the second matrix is non-singular since $t$ is non-singular.

By compairing (\[det\]) with the computations $[t e_i, t e_j]= t[e_i,e_j]$ we get the conditions $$t_{33}=\det A,\qquad B=\det(A) A, \qquad 
\left(\begin{array}{c}t_{43}\\ t_{53}\end{array}\right)=A \left(\begin{array}{c}t_{32}\\ -t_{31}\end{array}\right).$$

So for any $t\in \Aut({\mathfrak n }_{2,3})$, $(t_{ij})_{i,j}$ has the form $$\label{t}
\left(\begin{array}{ccc}
A &                                \begin{array}{cc}0\\0\end{array} & \begin{array}{cc}0&0\\0&0\end{array} \\
\begin{array}{cc} t_{31}&t_{32}\end{array} &      \det(A)                               