Show that $g$ is not topologically conjugate to the tent function Question: Define the function $g : [0, 1]\mapsto [0, 1]$ by

$g(x)=\begin{cases}3x&amp;\text{if}\;\;0\le x\le \frac{1}{3}\;\;\\{}\\
2-3x&amp;\text{if}\;\;\frac{1}{3}\le x\le \frac{2}{3}\;\;\\{}\\
3x-2&amp;\text{if}\;\;\frac{2}{3}\le x\le 1\;\;\\{}\\\end{cases}$

Show that $g$ is not topologically conjugate to the tent function.

Def.: The functions $f : X\mapsto X$ and $g : Y\mapsto Y$ (and the 
dynamical systems defined by them) are said to be topologically conjugate if 
there exists a homeomorphism $h : X\mapsto Y$ such that $g\circ h = h\circ f$. The 
function h is called a topological conjugacy between $f$ and $g$. 

If I must show that there is no $h(x)$ such that $g\circ h = h\circ f$, I don't have any clue how to guess some $h(x)$ at all; even though, it's not a proof it's 'many examples'. 

Also, I don't know which of tent functions writer is mentioning; i.e., for what $n$ in $T^n$. The pictures for $T^1$ and $T^2$ follows:

 

Thank you.
Consider the solutions of $g\circ h=\frac{1}{2}$ and $h\circ T=\frac{1}{2}$ (or $h\circ T^n=\frac{1}{2}$ if you want or need to prove it for any number of tents).

Solving $g\circ h=\frac{1}{2}$:

$g=\frac{1}{2}$ at $3$ points in the interior of $[0,1]$. These points are mapped by $h^{-1}$ to three interior points of $[0,1]$.

Solving $h\circ T=\frac{1}{2}$:

$h^{-1}$ maps $\frac{1}{2}$ to an interior point $a\in[0,1]$. Then $T=a$ (or $T^n$) at an even number of points.
Consider the solutions of $g\circ h=\frac{1}{2}$ and $h\circ T=\frac{1}{2}$ (or $h\circ T^n=\frac{1}{2}$ if you want or need to prove it for any number of tents).

Solving $g\circ h=\frac{1}{2}$:

$g=\frac{1}{2}$ at $3$ points in the interior of $[0,1]$. These points are mapped by $h^{-1}$ to three interior points of $[0,1]$.

Solving $h\circ T=\frac{1}{2}$:

$h^{-1}$ maps $\frac{1}{2}$ to an interior point $a\in[0,1]$. Then $T=a$ (or $T^n$) at an even number of points.