I think that if two Lie algebras are isomorphic, then their centers should be isomorphic - is this true? I am sure the answer is obvious to those in the know! Here is my attempt at a proof which looks fairly solid to me. Any pointers or thoughts would be appreciated.
Proposition: Isomorphic Lie algebras have isomoprhic centers.
Proof: Let $\mathfrak{L}_{1}, \mathfrak{L}_{2}$ denote two Lie algebras with $\mathfrak{L}_{1} \cong \mathfrak{L}_{2}$. That is, there exists a bijective algebra homomorphism $\phi_{1}:\mathfrak{L}_{1} \rightarrow \mathfrak{L}_{2}$ with unique inverse map $\phi_{2} = \phi_{1}^{-1}:\mathfrak{L}_{2} \rightarrow \mathfrak{L}_{1}$ such that
$$ \phi_{i}([a,b]) = [\phi_{i}(a),\phi_{i}(b)] \quad \forall \; a,b \in \mathfrak{L}_{i} \quad (i=1,2). $$
Now, let $a_{i},b_{i}\in \mathfrak{L}_{i}$ with $a_{i}=\phi_{j}(a_{j}), b_{i}=\phi_{j}(b_{j})$ for $a_{j},b_{j}\in Z(\mathfrak{L}_{j})\;$ ($j=1,2$ and $i\neq j$). Then
$$[a_{i},b_{i}] = [\phi_{j}(a_{j}),\phi_{j}(b_{j})] = \phi_{j}([a_{j},b_{j}]) = \phi(0) = 0$$
since $[a_{j},b_{j}]=0$ as both elements lie in the center of $\mathfrak{L}_{j}$. Hence we have shown that $a_{i},b_{i}\in Z(\mathfrak{L}_{i})$. That is, if $a_{1},b_{1}\in Z(\mathfrak{L}_{1})$, then $a_{2}=\phi_{1}(a_{1}), b_{2}=\phi_{1}(b_{1}) \in Z(\mathfrak{L}_{2})$ and if $a_{2},b_{2}\in Z(\mathfrak{L}_{2})$, then $a_{1}=\phi_{2}(a_{2}), b_{1}=\phi_{2}(b_{2}) \in Z(\mathfrak{L}_{1})$. Hence $Z(\mathfrak{L}_{1})\cong Z(\mathfrak{L}_{2})$. $\, \blacksquare$
I would prove it by noting that $[a,b]=0$ if and only if $[\phi(a),\phi(b)]=0$, so $[a,b]=0$ for all $b$ if and only if $[\phi(a),c]=0$ for all $c$ since $\phi$ is surjective.