Does equality of the complexifications imply equality of the algebras? Let $\mathfrak{g}$ and $\mathfrak{h}$ be two real Lie algebras. Now suppose their complexifications are isomorphic, that is, $$\mathfrak{g}_{\mathbb{C}}\simeq\mathfrak{h}_{\mathbb{C}}.$$.
Can I say anything about the "isomorphicness" of $\mathfrak{g}$ and $\mathfrak{h}$? i.e. do I necessarily have $$\mathfrak{g}\simeq\mathfrak{h}$$
Specifically I am curious about whether $$\mathfrak{su}(4)_{\mathbb{C}}=\mathfrak{sl}(4,\mathbb{C})\simeq \mathfrak{so}(6,\mathbb{C})=\mathfrak{so}(6,\mathbb{R})_\mathbb{C}$$ implies $$\mathfrak{su}(4)\simeq \mathfrak{so}(6,\mathbb{R})$$

A simpler counterexample in dimension $3$ is $\mathfrak{g}=\mathfrak{so}_3(\mathbb{R})$ and $\mathfrak{h}=\mathfrak{sl}_2(\mathbb{R})$. We have $$\mathfrak{g}_{\mathbb{C}}\cong\mathfrak{h}_{\mathbb{C}}\cong \mathfrak{sl}_2(\mathbb{C}),$$
but $\mathfrak{g}$ and $\mathfrak{h}$ are not isomorphic as real Lie algebras, because one algebra has a $2$-dimensional subalgebra, the other one does not.
If we furthermore assume that the real Lie algebras are simple, then we can use the machinery of the real classification, i.e., that we have one compact real form and one split form etc. The classification result in dimension $3$ for the simple real Lie algebras then is exactly $\mathfrak{so}_3(\mathbb{R})$ and $\mathfrak{sl}_2(\mathbb{R})$.
