If $f:X \to X$ is a continuous bijection and every point has finite orbit, is $f^{-1}$ continuous? If $f:X \to X$ (codomain and domain have the same topology) is a continuous bijection and every point has finite orbit, is $f^{-1}$ continuous? 
Note that the orbit being finite and $f$ being a bijection means for all $x$ means for all $x$ there is an $n>0$ such that $f^n(x)=x$.
I asked myself this question while answering another question on this site and ended up not getting anywhere. I suspect it should come from very basic facts or is false in general. I am leaning towards false at the moment but have not been able to construct a counter example.
One attempt I tried was to see if considering $X_F$, the set $X$ final topology with respect to $f,f^{-1}$ had the same topology, and I believe I got $f:X_F \to X$ was continuous but, could never get the same for $f^{-1}$.
This question is similar, without the finite orbit restriction.
 A: This seems to be a counterexample. Take $X = \mathbb{N}^2 \cup \{\infty\}$ with the metric where $\rho((m,n), \infty) = 1/n$ and 
$\rho((m,n), (p, q)) = 1/n + 1/q$ when $(m,n) \ne (p,q)$.
Define $f(\infty) = \infty$ and
$$
f((m,n)) = \cases{
   (m, n-1) & if $2 \le n \le m$, \\
   (m, m)   & if $n=1$, \\
   (m, n)   & otherwise.
}
$$
For every $m \in \mathbb{N}$ the set $\{ (m, 1), \dots, (m, m) \}$ is
an finite orbit and any point not in such a set is fixed.
Since $(\mathbb{N}^2, \rho)$ is discrete, we only need to consider
continuity at $\infty$. It is easy to verify that $\rho(f(m,n), \infty) \le 2\rho((m,n), \infty)$, so $f$ is continuous. On the other hand, every 
neighbourhood of $\infty$ contains a point $(m,m)$ for some value of $m$ and
$\rho(f^{-1}(m,m), \infty) = 1$, hence $f^{-1}$ is not continuous.

For a slightly more natural counterexample, you could construct a vector
space automorphism of $\mathbb{R}^\infty$ in a similar way, for example
such that
$$\begin{eqnarray}
Te_1 &=& 2e_2, Te_2 = \frac{e_1}{2}, \\
Te_3 &=& 2e_4, Te_4 = 2e_5, Te_5 = \frac{e_3}{4} \\
\text{etc.}&&
\end{eqnarray}
$$
Clearly $T$ is a bounded linear operator w.r.t. the Euclidean norm but its
inverse is not bounded. To see that it has finite orbits, note that the
orbits of basis vectors are finite, say $T^{k_i} e_i = e_i$, and it follows
for $x = \sum_{i=1}^n c_ie_i$ that $T^K x = x$, where $K = \prod_{i=1}^n k_i$.
