It is an old result that every $6$-connected graph is rigid in $\mathbb{R}^2$:

Lovász, László, and Yechiam Yemini. "On generic rigidity in the plane."

SIAM Journal on Algebraic Discrete Methods3, no. 1 (1982): 91-98. DOI.

It is natural to hope that sufficiently high connectivity implies rigidity in $\mathbb{R}^3$.

Is it known that there is some $k$ such that any $k$-connected graph is generically rigid in $\mathbb{R}^3$?Q.

Informally, $G$ is rigid if the distances between vertices connected in $G$
determine all the distances between vertices not connected in $G$.
More formally, $G$ is *generically rigid* in $\mathbb{R}^d$ if every generic representation in $\mathbb{R}^d$
is infinitesimally rigid—no
infinitesimal length-preserving velocities (if they exist) can be extended.
An embedded graph representation (a *framework*) is *generic* if the coordinates of its configuration do not satisfy
any non-trivial algebraic equation with rational coefficients.

There have been recent advances in 3D rigidity and I am unclear on the current status
of this question * Q*.

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