$\newcommand{\norm}[1]{\lVert{}#1\rVert{}}$ $\newcommand{\Id}{\textrm{Id}}$ $\newcommand{\Fix}{\textrm{Fix}}$
Let $(E, \norm{\cdot})$ be a complete metric space, $K$ a convex compact of $E$.
For all $1$-lipschitz continuous map $f : K \to K$, then $f$ has a fix point (if we consider $a \in K$ and $\varphi_n : x \mapsto \dfrac{a}{n} + \dfrac{n + 1}{n} f(x)$ defined over $K$, $\varphi_n$ is $k$-lipschitz continuous with $k < 1$ so that we can show it has a fixed point, then get back to $f$).
For example, $\Id_{K}$ denotes the identity map from $K$ to $K$, such map is of course $1$-lipschitz continuous (e.g. $\forall (x, y) \in K^2, \norm{\Id_{K}(x) - \Id_{K}(y)} \leq \norm{x - y}$ — in fact it is equal.)
If we denote finally $\Fix(f)$ the set of the fixed points of $f$.
For example, $\Fix(\Id_{K}) = K$.
In which cases, if we take a map $f : K \to K$ $1$-lipschitz continuous (such that there is at least one fix point), do we have $\Fix(f) = \{ x_0 \}, x_0 \in K$ ?
I conjecture this is true for all maps distinct from $\Id_{K}$, but unsure how to take this problem.
Let $K=[0,2]$ and define $f:K\to K$ by $f(x)=x$ for $x\leq 1$ and $f(x)=1$ if $x>1$. Then $f$ is $1$-Lipschitz continuous and the set of fixed points of $f$ is $[0,1]$.