Let $\mathfrak g:=(V,d,[\cdot,\cdot])$ be a differential graded Lie algebra (DGLA) where $d$ is the zero differential. 


Intrinsic formality: The DGLA $\mathfrak g$ will be said intrinsically formal if any $L\_\infty$-algebra $l$ admitting $\mathfrak g$ in cohomology (i.e. $H(l)=\mathfrak g$) is formal (i.e. there exists a $L\_\infty$-quasi isomorphism $\mathfrak g\to l$). 


Rigidity: The DGLA $\mathfrak g$ will be said rigid if any $L\_\infty$-algebra $(V,l')$ such that $l'\_1=0$, $l'\_2=[\cdot,\cdot]$ is isomorphic to $\mathfrak g$. In other words, $\mathfrak g$ does not admit non-trivial deformations as a $L\_\infty$-algebra. 


**Question 1**: What is the relation (equivalence, implication) between these two notions?


Let me denote $NR(\mathfrak g)$ the DGLA controlling the deformations of $\mathfrak g$ as a $L\_\infty$-algebra and let $H^1(NR(\mathfrak g))$ be the associated first cohomology group. 


**Question 2**: Does any of the previous notions entails the vanishing of $H^1(NR(\mathfrak g))$, or conversely ?



Intrinsic formality versus rigidity of a differential graded Lie algebra
As I said in the comments, the answer to question 1 is that the two notions are equivalent (at least in characteristic zero, which I will assume throughout). 


Assume $\mathfrak{g}=(V,0,[\cdot,\cdot])$ is intrinsically formal. Now let $(V,l')=(V,0,l'\_2=[\cdot,\cdot],l'\_3,\ldots)$ be an $L\_\infty$ algebra as in the definition of rigidity. Then the homology of $(V,0)$ is $(V,0)$ and so the induced truncated Lie algebra on $H(V,l')$ is precisely $\mathfrak{g}$. By intrinsic formality, there exists an $L\_\infty$ quasi-isomorphism from $\mathfrak{g}$ to $(V,l')$. Any $L\_\infty$ quasi-isomorphism between two $L\_\infty$ algebras with zero differential is in fact an $L\_\infty$ isomorphism, so $(V,l')$ is isomorphic to $\mathfrak{g}$. Thus $\mathfrak{g}$ is rigid.


On the other hand, assume $\mathfrak{g}$ is rigid and let $(W,l)$ be an $L\_\infty$ algebra with 
$$H(W)^{\text{trunc}}:=(H(W),0,(l\_2)\_\*)$$ isomorphic to $\mathfrak{g}$ as in the definition of intrinsic formality. The homological perturbation lemma says that there exists an $L\_\infty$ algebra $$H(W)^{\text{trans}}:=(H(W),0,(l\_2)\_\*,l\_3^{\text{trans}},\ldots)$$
equipped with a quasi-isomorphism to $(W,l)$. Note that the truncation of $H(W)^{\text{trans}}$ is $H(W)^{\text{trunc}}\cong \mathfrak{g}$. Then by rigidity, $H(W)^{\text{trans}}$ is isomorphic to $\mathfrak{g}$. Composing this with the quasi-isomorphism $H(W)^{\text{trans}}\to (W,l)$ gives a quasi-isomorphism so that $\mathfrak{g}$ is intrinsically formal.


