The closure of all periodic homeomorphisms of circle Let $X$  be the space of  all continuous  maps  from $S^{1}$ to $S^{1}$. $X$ is  equipped with $C^{0}$ topology. Let $Y\subset X$ be the union of all  finite order homeomorphisms i.e:  all homeomorphisms $\phi$ with $\phi^{n}=Id$ for  some natural number $n$. Assume that $g\in X$ belongs to $\bar{Y}-Y$. Some  questions  about such $g$:
1.Is $g$ necessarilly a  homeomorphism?If yes, is it necessarilly order preserving?


*Is it true to say that $g$ satisfies at least one of the following conditions:


I: The periodic points of  $g$ is  dense
II: $g$  has  a  dense orbit 
More  generally, to what extent, the elements of $\bar{Y}-Y$ are known, dynamically?
What  about if  we replace $S^{1}$ with another manifold and study the elements of $\bar{Y}-Y$? What  about if we consider $C^{k}$ topology for $k>0$?
Note: A possible operator theoretical version for  question $1$ could be the following:Let $A$ be a  $C^{*}$ algebra. Let $X$ be the space of all unital morphisms and $Y\subset X$ be the union of all finite order automorphisms. Assume that $T\in \bar{Y}-Y$. Is $T$ necessarilly an automorphisms.(Actually, is it surjective?)
 A: The answer to Question 2 is no. In fact, there is an orientation-preserving homeomorphism $g\colon \mathbb{R}/\mathbb{Z}\to\mathbb{R}/\mathbb{Z}$ such that $g\in\overline{Y}\setminus Y$ and such that it has exactly one fixed point (consequently, it has no dense orbit and it is not periodic):
For each $t\in\mathbb{R}$, let $\tilde g_t\colon\mathbb{R}\to\mathbb{R}$ be given by $\tilde g_t(x):= x + 1/8(1-\cos(2\pi x))+t$. Notice $\tilde g_t$ is the lift of a circle homeomorphism $g_t$. Let $g=g_0$ and observe $0\in\mathbb{R}/\mathbb{Z}$ is the only fixed point of $g$.
We want to show that $g\in\overline{Y}\setminus Y$. To do this, let $\rho_t=\rho(\tilde g_t)$ denote the rotation (also called translation) number of $\tilde g_t$. Observe that $\rho_t$ depends continuously on $t$ and 
$$ \rho_t > \rho_0=0, \quad\forall t>0.$$
In particular, this implies that there exists a strictly decreasing sequence $(t_n)$ with $t_n\to 0$ such that $\rho(t_n)$ is a Diophantine (irrational) number. Hence, by Herman-Yoccoz linearization theorem (see for instance https://eudml.org/doc/82144), for each $n$ there exists a $C^\infty$-diffeomorphism $h_n\colon\mathbb{R}/\mathbb{Z}\to\mathbb{R}/\mathbb{Z}$ such that $h_n\circ g_{t_n}\circ h_n^{-1}=R_{\rho_{t_n}}$, the rigid rotation of angle $\rho_{t_n}$. Then one can choose a sequence of sufficiently small positive real numbers $(\epsilon_n)$ such that $t_n+\epsilon_n\in\mathbb{Q}$ and 
$$h_n^{-1}\circ R_{t_n+\epsilon_n}\circ h_n\to g, \quad \text{in the } C^\infty\text{-topology}.$$
Finally, since $t_n+\epsilon_n\in\mathbb{Q}$, we have that $h_n^{-1}\circ R_{t_n+\epsilon_n}\circ h_n\in Y$ and consequently, $g\in\overline{Y}$. 
A: The answer to both questions 1 and 2 is false, due to the following example.
Consider $(n+1)$ consecutive intervals $I_0,...,I_n$ of lengths $1,2,4,...,2^{n-1}, 2^n$. Let the map $T_n$ cyclically permute them in an affine way. In fact, if you choose $I_j=[2^j,2^{j+1}]$, then 
$$
T_n(x)=\begin{cases}
2x, & x \not\in I_{{n}}\\
2^{-n}x, & x\in I_{n}.
\end{cases}
$$
Now, make an affine change of the coordinates so that the union $I_0\cup\dots\cup I_n=[1,2^{n+1}]$ becomes (always the same) circle $[0,1]/(0\sim 1)$. Then, the constructed maps on this circle $C^0$-converge, as $n\to\infty$, to
$$
T(x)=\begin{cases}
2x, & x \in [0,1/2]\\
0, & x\in [1/2,1].
\end{cases}
$$
This map is not a homeomorphism (it collapses the interval $[1/2,1]$ to a single point 0). Also, the orbit of any point falls to the fixed point 0 in a finite number of steps, so there is only one periodic point, and there is no dense orbit.
