(cf. The Classification of Three-dimensional Lie Algebras Theorem 3.2)
Let $L_{\theta\vartheta}$ be the Lie algebra with structure $$ (*)\qquad [x, y] = z, [x, z] = −ϑy,[y, z] = θx. $$ Let $<\theta,\vartheta,\theta\vartheta>$ denote the matrix representation of the Killing form of $L_{\theta,\vartheta}$, respect to the basis $\{x,y,z\}$.
Theorem.3.2. The Killing forms $<\theta,\vartheta,\theta\vartheta>$ and $<\alpha,\beta,\alpha\beta>$ are isometric if and only if $L_{\theta\vartheta}$ and $L_{\alpha\beta}$ are isomorphic.
Proof. If $L_{\theta\vartheta}$ and $L_{\alpha\beta}$ are isomorphic then an isomorphism $\phi$ preserves the Lie products, i.e., $\phi[u,v]=[\phi(u),\phi(v)]$ for all $u,v\in L_{\alpha\beta}$. It thus follows that $L_{\alpha\beta}$ and $L_{\theta\vartheta}$ will have the same adjoint matrices and hence the same Killing forms.
Question: I don't understand why they have the same adjoint matrices. The adjoint matrices should be depends on the basis which I choose, so are the Killing form matrices.
The notation $L_{\alpha,\beta}$ means that there already exists a basis $\{\hat x,\hat y,\hat z\}$ satisfies the structure $(*)$. If we represent the Killing form of $L_{\alpha,\beta}$ by the basis $\{\phi(x),\phi(y),\phi(z)\}$, then the Killing form I obtain should be an another matrix, rather than $<\alpha,\beta,\alpha\beta>$.
I've tried to use the property that
If $B$ is the Killing form of $L$ and $s$ is a automorphism of $L$, then $B(s(u),s(v))=B(u,v)$.
to link $\{\phi(x),\phi(y),\phi(z)\}$ and $\{\hat x,\hat y,\hat z\}$, but I found nothing.
We can use an Lie algebra isomorphism $f\colon L_1\rightarrow L_2$ to rewrite one Lie algebra, say $L_1$, so that its Lie brackets coincide with the ones of $L_2$. Then they have of course identical adjoint matrices. In a sense, it is trivial, that isomorphic Lie algebras share such invariant properties, because they are actually the same after just rewriting.