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Suppose $L$ is a nilpotent finite-dimensional Lie algebra over $\mathbb{Q}$ of class $c$. We can define an associated graded Lie algebra to $L$ that, as a vector space, is:

$$\bigoplus_{i=1}^c \gamma_i(L)/\gamma_{i+1}(L)$$

where $\gamma_i$ denotes the $i^{th}$ member of the lower central series, and where the Lie bracket is obtained from the Lie bracket of $L$ in a natural fashion (as follows: the Lie bracket of $L$ induces a map $\gamma_i(L) \times \gamma_j(L) \to \gamma_{i+j}(L)$ which descends to a bilinear map $\gamma_i(L)/\gamma_{i+1}(L) \times \gamma_j(L)/\gamma_{j+1}(L) \to \gamma_{i+j}(L)/\gamma_{i+j+1}(L)$. The Lie bracket in the associated graded is obtained from these constituent maps being summed up.

I believe that, in general, $L$ and its associated graded Lie algebra do not necessarily have to be isomorphic. However, I'm having a little trouble coming up with an explicit example where they're not. In particular: (i) if $L$ has class two, then it is isomorphic to its graded Lie algebra, (ii) if $L$ is a "free nilpotent" Lie algebra on a certain number of generators, then again it is isomorphic to its graded Lie algebra.

This has some relation with the Malcev Lie correspondence, but I preferred not to introduce the complication of nilpotent groups into the question. If viewed that way, the question is: for a rationally powered finitely generated nilpotent group, is the Malcev Lie ring the same as its associated graded Lie ring? Some of the discussion at Relationship between the cohomology of a group and the cohomology of its associated Lie algebra. may also be relevant.

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You can find an example on pages 151-152 of Y.Shalom, "Harmonic analysis, cohomology, and the large-scale geometry of amenable groups", Acta Math, vol. 192 (2004) 119-185.

In this example, the corresponding nilpotent groups are not even quasi-isometric to each other.

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The example Misha refers to is 7-dimensional.

The smallest examples are 5-dimensional, there are exactly two of them. The first $\mathfrak{g}$ is the Lie algebra with basis $(x_1,x_2,x_3,x_4,x_5)$ with nonzero brackets $[x_1,x_3]=x_4$, $[x_1,x_4]=[x_2,x_3]=x_5$. The associated Carnot graded Lie algebra $\mathfrak{g}'$ (often called "associated graded" but this is misleading) has the same basis with nonzero brackets $[x_1,x_3]=x_4$, $[x_1,x_4]=x_5$. They are not isomorphic (because $\mathfrak{g}$ has 1-dimensional center and $\mathfrak{g}'$ has 2-dimensional center (actually it splits as a direct product product of the subalgebras with basis $(x_1,x_3,x_4,x_5)$ and $(x_2)$).

The second example $\mathfrak{h}$ has basis $(y_1,y_2,y_3,y_4,y_5)$ with nonzero brackets $[y_1,y_i]=y_{i+1}$ for $i=2,3,4$ and $[y_2,y_3]=y_5$. The associated Carnot graded Lie algebra $\mathfrak{h}'$ has the same basis with nonzero brackets $[y_1,y_i]=y_{i+1}$ for $i=2,3,4$; they are not isomorphic, for instance because $\mathfrak{h}'$ has an abelian 1-codimensional ideal unlike $\mathfrak{h}$.

In the first case, although the Betti numbers are the same, the groups are not quasi-isometric because the real cohomology algebras are not isomorphic as graded real algebras, and we can apply Sauer invariance theorem (which improves Shalom's).

In the second case, it can be checked that the real cohomology algebras are isomorphic as graded algebras, and it is not known whether the nilpotent groups are quasi-isometric, it's indeed the smallest open case.

In dimension 6 and beyond, there are plenty of examples.

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