Is every 1-million-connected graph rigid in 3D? 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 Methods 3, no. 1 (1982): 91-98.
DOI.

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

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

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.
 A: update: this is an answer without generic configuration assumption
I am afraid that no. Take two half-planes with a common boundary line $a$, and many points both on $a$ and in these half-planes. Join by edges all pairs of points in the same half-plane. Now rotate one of the half-planes around $a$.
A: I think this is still an open problem, but recent work of Clinch, Jackson, and Tanigawa (almost) shows every $12$-connected graph is generically rigid in $\mathbb{R}^3$.
In that paper, they prove that $12$-connectivity is sufficient to force rigidity in the $C_2^1$-cofactor matroid (see the paper for precise definitions). In an earlier paper, the same authors showed that the $C_2^1$-cofactor matroid is the unique maximal abstract $3$-rigidity matroid.  A long-standing conjecture in rigidity theory is that the unique maximal abstract $3$-rigidity matroid is in fact the generic $3$-dimensional rigidity matroid. If you believe this conjecture, then the answer to your quesiton is yes, with one million replaced with $12$.
Acknowledgement. This answer is entirely due to Katie Clinch.
