Random walk on $\mathbb Z/m\mathbb Z$ converges to uniform distribution 
Let $(X_t)_t$ be the standard continuous time random walk on $\mathbb Z/m\mathbb Z$ with $X(0)=[0]$ almost surely, then I want to show that for $t \rightarrow \infty$, $\lim_{t \rightarrow \infty} P(X(t)=[n]\mid X(0)=[0])= \frac{1}{m}$ for any $[n] \in Z/mZ.$ 

More precisely the random walk is defined in such a way that the rate of going from $[n]$ to $[n+1]$ is 1 and also the rate of going from $[n]$ to $[n-1].$
By rate I mean the number $$\left.\frac{d}{dt}P(X(t)=[n+1]\mid X(0)=[n])\right|_{t=0}.$$
The problem here is that $m$ is not fixed, so we cannot explicitly constuct the transition probabilities, I guess. Nevertheless, I guess that the simple structure of the random walk enables us to get this result still somehow.
 A: There are two approaches. One way is to use the procedure for continuous-time markov chain analysis by finding an embedded markov chain which operates in discrete time. You can read details here. In this case, the embedded chain is simply $X_n$ which is a simple random walk on $\mathbb{Z}/m\mathbb{Z}$. This is a finite state space, and $X_n$ is aperiodic for $m$ odd, where it has a unique stationary distribution. You can show this stationary distribution $\pi$ is $\pi_t=1/m$ in a number of ways. The first is to write down the transition matrix $P$ for $X_n$, and verify that the above stationary distribution satisfies $\pi P=\pi$, and then invoke uniqueness. An easier way is to notice that $X_n$'s distribution is cyclically invariant under rotations of $\mathbb{Z}/m\mathbb{Z}$ which means that $\pi$ must have equal probability 
for all states.
To carry this over to a continuous time markov chain, you need the following result which connects to the embedded markov chain:
$$\lim_{t\rightarrow \infty}P(X_t = j | X_0=i) = \frac{\pi_j/v_j}{\sum_k \pi_k/v_k},$$
where $v_k$ are the transition rates for your continuous time chain, all of which are $1$.
When $m$ is even $X_n$ is periodic. Despite this $X(t)$ will still have a stationary distribution, primarily because periodicity does not exist in continuous time. To see this you can use the fact that if a probability vector $\pi$ satisfies the local balance equations (see page 253 of this):
$$\pi_iq_{ij}=\pi_jq_{ji},$$
where $q_{ij}$ are transition rates, then $\pi$ must be a stationary distribution of $P(t)$. You can double check $\pi_i=1/m$ satisfies the above equations with $q_{ij}=1$ for $|i-j|=1$ and 0 otherwise. Note that not all stationary distributions satisfy local balance equations. 
This reasoning uses the fact that if $\pi$ is the stationary distribution of a continuous time Markov chain, it is unique.
