Let $n>0$, $\mathbb S^n$ be $n$-sphere and $1\in \mathbb S^n$ be its north pole.

A am looking for an example of compact manifold $M$ with a continuous $n$-parameter family off maps $h_x\colon M\to M$, $x\in \mathbb S^n$ such that

- $h_1=\mathop{\rm id}_M$
- For any Riemannian metric $g$ on $M$ and any family of maps $h'_x\sim h_x$ (i.e., such that the maps $h,h'\colon \mathbb S^n\times M\to M$ are homotopic) there is $x\in\mathbb S^n$ such that the Lipschitz constant of $h'_x\colon (M,g)\to(M,g)$ is $\ge 10000$.

**Comments.**

- I think about this problem for few years; I do not think it is an easy one, but I want to make sure that I did not miss a well known trick. A positive answer might have interesting consequences for collapsing with lower curvature bound.
- If $n=0$, so $\mathbb S^0=\{+1,-1\}$ then $M=\mathbb T^k$ and $h_{-1}=$ a hyperbolic linear map with a big eigenvalue will do.
- Denote by $F$ the space of all maps $M\to M$. If $\pi_n(F,\mathop{\rm id}_M)$ is finitely generated then for every $h_x\colon (M,g)\to (M,g)$ there is $h'_x$ with a fixed Lipschitz constant $L=L(M,g)$ for all $x$.
- I know few examples of families $h_x$ for which one can not find a metric $g$ and homotopic family $h'_x$ with Lipschitz constant $=1$ (in other words the family can not be made isometric).
- I do not know an answer even if instead of 10000 I would have $1+\varepsilon$ for a fixed $\varepsilon> 0$.