I have two questions and I am very confused about the concepts
Can a Markov process of order one also be a a Martingale? Is any Markov process of order one also a Martingale? For 1. I would say yes, since a Markov of order one is also an AR(1) it would be a martingale if c=0 and phi=1. However, I am very confused about the second. My answer for 2 would be no, but I cannot explain it properly.
Can anyone maybe help me?
On
A first-order Markov chain can be a martingale, but need not be. In order for it to be a martingale, you need $\mathbb E[X_{n+1} | X_n] = X_n$. So for example a simple unbiased random walk (equal probability of steps of $+1$ and $-1$) is a martingale. If you want it to be a finite Markov chain on states $a, \ldots, b$, make $a$ and $b$ absorbing. On the other hand, a biased random walk is not a martingale.
The answer by @RobertIsrael shows that the Markov property does not imply the martingale property, even when the Markov chain has order one.
As for the relevance to autoregressive processes:
An AR(1) process $(X_t)$ (where $t$ is a discrete time index) satisfies $$ X_{t+1} = c + \phi X_t + \varepsilon_{t+1}\tag1 $$ It follows that every AR(1) process is Markov of order one, since the distribution of $X_{t+1}$ given $X_1,X_2,\ldots ,X_t$ is the same as the distribution of $X_{t+1}$ given $X_t$. Taking conditional expectations of (1) given $X_t$, we get: $$ E(X_{t+1} \mid X_t) = c + \phi X_t\tag2 $$ This tells us that an AR(1) process is not a martingale except in the special case when $c=0$ and $\phi=1$.