Continuity of a map with constrains Let  $A_i$ be a disjoint union of finite number of closed sub intervals of $[0,1]$, $1\leq i\leq n$. Each of $A_i$ has non-empty intersection with $A_j$. However, the intersection of each triple of $A_i$'s is empty. Let $h$ be a map from $[0,1]$ to $[0,1]$, such that for all $i$ the map $h$ sends $A_i$ into its complement. Is it possible to find $A_i$ and $h$, so that $h$ is homeomorphism? I know how to do that for $n=1,2,3$.
EDIT, stronger related question:
Let  $A_i$ be a family of sets of the circle $S^1$, so that:
1) Each $A_i$ is a union of open intervals 
2) for all indices i, j, intersection of $A_i$ and $A_j$ is not empty,
3) for any point p in $S^1$, there are at most two indices, i and j, say, so that p is in $A_i$ and $A_j$.
4) There is h a homeomorphism of $S^1$ so that for all indices i, $hA_i$ contains the complement of $A_i$,
 A: It seems the following.
The answer is trivial, in some cases even if we fix the collection of sets $A_i$. Put $A=\bigcup A_i$. 
If $A=I$ then each homeomorphism $h:[0,1]\to [0,1]$ has a fixed point $a$. Then $a\in A_i$ for some $i$. Thus $h(a)=a\in A_i$, which contradicts to condition $A_i\cap h(A_i)=\emptyset$.
If $A\not\ni 0$ then since $A$ is compact, there exists a number $\varepsilon\in (0,1)$ such that $[0,\varepsilon]\cap A=\emptyset$. Then we can easily define a homeomorphism $h:[0,1]\to [0,1]$ such that $h(0)=1$, $h(\varepsilon)=\varepsilon$, $h(1)=0$, and $h$ is affine on each of the segments $[0,\varepsilon]$ and $[\varepsilon,1]$. Then $h(A)\cap A=\emptyset$ and hence 
$h(A_i)\cap A_i=\emptyset$ for each $i$.
EDIT:
The edited question maybe even simpler, provided instead of Condition 4 "$h(A_i)\supset S^1\setminus A_i$" we have a condition "$h(A_i)\cap A_i=\emptyset$".  Such  the system of objects exists. Let $\varepsilon>0$ be a sufficiently small number. 
Indentify the circle $S^1$ with the set $\{z\in\mathbb C:|z|=1\}$. 
For each $j\in\{1,2,3\}$ put $A_j=\{z\in S^1: -\varepsilon+2\pi (j-1)/3<\arg z<2\pi j/3+\varepsilon\}$. Put $h(z)=ze^{i\pi}$ – the rotation on 180 degrees. Then $h(A_j)\cap A_j=\emptyset$ for each $j$.
EDIT 2:
Consider the (standard normed) measure $\mu$ on the circle $S^1$ and for the sake of simplicity we assume that the circle $S^1$ has the measure 1. For each $1\le i\le n$ let $\mu_i=\mu(A_i)$ and $\nu_i=\mu(h(A_i))$. Condition 4 implies that $A_i\cup h(A_i)=S^1$. The following two cases are possible. If the set $A_i$ is the whole circle $S^1$ then $h(A_i)=S^1$. If $A_i\ne S^1$ then since the circle $S^1$ is connected, it cannot be presented as a union of two its disjoint non-empty open subsets, therefore the intersection $A_i\cap h(A_i)$. is an open nonempty set, which contains an arc of non-zero length, so $\mu(A_i\cap h(A_i))>0$. In both cases we have $\mu_i+\nu_i>1$. From the other side, Condition 3 and Dirichlet principle imply that $\sum_i\mu_i\le 2$ (and similarly $\sum_i\nu_i\le 2$). So $n<\sum_i\mu_i+\nu_i\le 4$.   
From the other side, it seems that we can easily construct the system satisfying Conditions 1-4 for each $n\le 3$. For instance, for $n=3$ we simply put $\varepsilon=\pi/3$  in the construction from EDIT. 
A: Here is my reading of your question, please, correct me if I am wrong: 
Question. Given a collection of subsets $\{A_i, i=1,...,n\}$ satisfying the hypothesis appearing in your question, find an index $i$ and a continuous map $h: [0,1]\to [0,1]$, so that $h(A_i)\subset A_i^c:=[0,1]\setminus A_i$. 
Answer: 


*

*If $n=2, A_1=A_2=[0,1]$, such $i$ and $h$ do not exist, since $A_i^c=\emptyset, i=1,2$. 

*Assume now that $A_i\ne [0,1]$ for some $i$, this is automatic for $n>2$ because of the triple intersection condition. Then, pick index $i$ so that $A_i\ne [0,1]$ and a point $a\in A_i^c$ and let $h: [0,1]\to \{a\}\subset [0,1]$ be the constant map. 

*You may want to add the extra assumption that $h: [0,1]\to [0,1]$ is surjective. Then the above construction can be modified as follows. Define $h|A_i$ to be constant as above; pick a closed subinterval $J\subset [0,1]$ disjoint from $A_i$ and let $h: J\to [0,1]$ be a continuous surjective map. On the complementary intervals of $[0,1]\setminus (A_i \cup J)$, extend $h$ by linearity. The result is a continuous map satisfying the requirements. 
There is one more question one can ask along these lines: Find a collection of subsets $A_i$ (satisfying your requirements as before) and a continuous map $h: [0,1]\to [0,1]$ so that for all $i=1,...,n$, we have $h(A_i)\subset A_i^c$. If this is what you had in mind, you should correct your question and I will think about it. 
