Suppose we are given a random variable $\sigma : \Omega \to S_n$ on the symmetric group of $n$ letters and we are given numbers representing the probabilities
$$p_{k,l} = \mathbb P(\sigma(k) = l).$$
Ignore for now that they are defined via the distribution of $\sigma$ and just consider numbers $p_{k,l} \geq 0$ with $\sum_{k=1}^n p_{k,l} = \sum_{l=1}^n p_{k,l} = 1$.
Does there always exist a distribution, i.e. probabilities $\mathbb P(\sigma = \tau)$ summing to $1$, such that $p_{k,l} = \mathbb P(\sigma(k) = l)$? Is there a canonical choice (maybe with maximal support)?
Of course, we have
$$\mathbb P(\sigma = \tau) = \mathbb P(\sigma(1) = \tau(1),\dots, \sigma(n) = \tau(n)),$$
but the joint distribution of $(\sigma (1),\dots, \sigma(n))$ is not known and they cannot be considered as independent.
We can write
$$\mathbb P(\sigma = \tau) = \mathbb P(\sigma(1) = \tau(1)|\sigma = \tau\text{ on } \{2,\dots, N\}) \mathbb P(\sigma(2) = \tau(3)|\sigma = \tau\text{ on } \{3,\dots, N\})\cdots \mathbb P(\sigma_N = \tau_N),$$
and this seems to be in the "permutation spirit", but I still don't see what probabilities one should use here.
Note that we are given $N^2$ numbers, but need to determine $N!$ numbers. The distribution of $\sigma$ cannot be unique in general as the following example shows:
Let $G$ be a transitive subgroup of $S_n$ and suppose $\sigma$ has uniform distribution on $G$. Then
$$\mathbb P(\sigma(k) = l) = \sum_{\tau \in G} \mathbb P(\sigma(k) = l|\sigma = \tau) \mathbb P(\sigma = \tau) = \frac{1}{|G|} |\{\tau \in G : \tau(l) = j\}|.$$
By the transitivity there exists a $\pi\in G$, such that $\pi(j) = l$. So
$$|\{\tau \in G : \tau(l) = j\}| = |\{\tau \in G : \pi\circ \tau(l) = \pi(l)\}| = |\{\tau \in G : \tau(l) = l\}| = |\operatorname{Stb}(l)|,$$
where $\operatorname{Stb}(l)$ is the stabilizer of $l$ under the action of $G$ on $\{1,\dots, n\}$. Since the action is transitive and by the orbit-stabilizer theorem we have
$$|\operatorname{Stb}(l)| = \frac{|G|}{N}.$$
Therefore,
$$\mathbb P(\sigma(k) = l) = \frac1N.$$
Notably, this does not depend on $G$. By choosing $G = S_n$ and $G = A_n$ (alternating group) we get two different distributions for $\sigma$ with the same probabilities $\mathbb P(\sigma(k) = l)$.
Best Answer
Yes, a distribution always exists. The matrix $P:=(p_{k,\ell})_{k,\ell=1}^n$ is doubly stochastic, so the Birkhoff–von Neumann theorem implies that $P$is a convex combination of permutation matrices. That is, $P=\sum_{i=1}^m a_i Q_i$, where $a_i\ge 0$, $\sum_{i=1}^m a_i=1$, and each $Q_i$ is a permutation matrix. This defines a distribution on $S_n$, where $P(\pi_i)=a_i$ for the permutation $\pi_i$ associated to $Q_i$.
I cannot say anything about a canonical choice of distribution.