Definition Let $M$ be a subset of a metric space $X$ and $T, I :M\to M$ be $M$-invariant maps. Then the pair $(T,I)$ is called compatible if $$\lim_{n\to\infty}d(ITx_n, TIx_n)=0$$ whenever ${x_n}$ is a sequence such that $\lim_{n\to\infty}Ix_n=\lim_{n\to\infty}Tx_n=t$ for some $t\in M$.
Example Let $X=\mathbb{R}$ with the usual metric and $M=[0,1]$. Let $T, I :M\to M$ be defined by $$ T(x) = \begin{cases} \frac{1}{2} & \text{if $0\leq x\leq \frac{1}{2} $} \\ \frac{x}{2}+\frac{1}{4} & \text{ if $\frac{1}{2}\leq x\leq 1$ } \end{cases}$$ and $$ I(x) = \begin{cases} \frac{1}{2} & \text{if $0\leq x\leq \frac{1}{2} $} \\ 1-x & \text{ if $\frac{1}{2}\leq x\leq 1$ } \end{cases}$$ Question Are $I$ and $T$ compatible?
Notice that $I$ and $T$ are continuous. If $(x_n)\subseteq [0,1]$ satisfies $\lim Ix_n=\lim Tx_n=t$ for some $t$, then $t$ lies in the closure of both $I(M)=[0,1/2]$ and $T(M)=[1/2,1]$, so $t=1/2$. Using the continuity of $T$ and $I$, we obtain $$\lim TIx_n=T(1/2)=1/2=I(1/2)=\lim ITx_n,$$ and this implies that $T$ and $I$ are compatible.