Markov chains with continuous time - is $P(2)$ having a given form even possible? 
Is it possible that
$$
\begin{vmatrix}
0 & 1 \\
1 & 0
\end{vmatrix} = P(2)
$$
for any Markov semigroup $\{P(t), t \geq 0\}$?


Recalling what properties must hold for transition matrix $P$:

*

*$\forall t>0$ every row of matrix must sum to 1 and all entries must be non negative

*$P(0) = I$ and $\lim_{t \to 0^+} P(t) = I$

*$P(s+t) = P(s)P(t)$
Any hints appreciated. I suppose we're looking for some kind of contradiction but I couldn't find any clues.

Let us abbreviate
$$
  A :=
  \begin{pmatrix}
    0 & 1 \\
    1 & 0
  \end{pmatrix}.
$$
Claim: A Markov semigroup $(P(t))_{t \in [0,\infty)}$ with $P(2) = A$ does not exist.
Here are three different proofs, so you can choose your favourite one. :-)
Proof 1 ($A$ has no real square root).
There is no matrix $B \in \mathbb{R}^{2 \times 2}$ such that $B^2 = A$; indeed, the equality $B^2 = A$ would imply $-1 = \det(A) = (\det(B))^2 \ge 0$.
Proof 2 (Perron-Frobenius theory).
For Markov semigroup $(P(t))_{t \in [0,\infty)}$ on $\mathbb{R}^n$ it follows from Perron-Frobenius theory that each matrix $P(t)$ has only one eigenvalue on the complex unit circle - namely the number $1$.
However, the matrix $A$ has the eigenvalues $1$ and $-1$.
Proof 3 (The isometry group with respect to the infinity norm is discrete).
Endow $\mathbb{R}^n$ with the infinity norm and assume that $(P(t))_{t \in [0,\infty)}$ is a Markov semigroup on $\mathbb{R}^n$ such that $P(2)$ is a permutation matrix. We show that this implies $P(t) = I$ (:= identity matrix) for all $t \ge 0$:
Each $P(t)$ acts contractively with respect to the infinity norm on $\mathbb{R}^n$ (via $x \mapsto P(t)x$). But $P(2)$ is an isometry with respect to this norm, which readily implies that each $P(t)$ is an isometry. Hence, each $P(t)$ is a signed permutation matrix (see this MathOverflow post). Since every entry of each $P(t)$ is $\ge 0$, this implies that each $P(t)$ is a permutation matrix.
From $P(t) \to I$ as $t \downarrow 0$ we now conclude that $P(t) = I$ for all sufficiently small times $t$. The semigroup law then implies that $P(t) = I$ for all times $t$.
In particular, the permutation matrix $P(2)$ must be equal to $I$, too.
Remark. Proofs 2 and 3 are, admittedly, quite an overkill if you are merely interested in the concrete example from the question. But on the other hand, proofs 2 and 3 give us more structural insight and work in more general situations than proof 1.
