It is known that if $X$ is a compact manifold, then for every $\epsilon>0$, there is $\delta>0$ such that for every finite sequence $\{(x_i, y_i)\}_{i=0}^n\subset X$ with $x_i\neq y_i$ and $d(x_i, y_i)<\delta$, there is a homeomorphim $h:X\to X$ such that $d(h(x), x)<\epsilon$ for all $x\in X$ and $h(x_i)=y_i$ for all $0\leq i\leq n$.
Question. Above property is hold on Banach spaces? this means that
if $X$ is a Banach space, Can I say that for every $\epsilon>0$, there is $\delta>0$ such that for every finite sequence $\{(x_i, y_i)\}_{i=0}^n\subset X$ with $x_i\neq y_i$ and $||x_i-y_i||<\delta$, there is a operator $h:X\to X$ such that $||h(x)-x||<\epsilon$ for all $x\in X$ and $h(x_i)=y_i$ for all $0\leq i\leq n$.
Surely not, if you mean bounded linear map when you say operator. First note that (as commented by Mindlack) $\|h(x)-x\|<\epsilon$ for all $x$ means $h$ is the identity so you should replace it with $\|h(x)-x\|\leq \epsilon \|x\|$ for all $x$. Even then the result is false. Consider $\mathbb R$ as a one-dimensional Banach space. In this case $h(x)=cx$ for some real number $c$. If $|x_i-y_i|<\delta$ you want $cx_i=y_i$ for all $i$ which can hold only when $\frac {y_i} {x_i}$ is same for all $i$.