In this question, a stochastic square matrix is a real square matrix where all the rows sum up to $1$ and all the entries are between $0$ and $1$. Permutation matrices are examples of stochastic square matrices, for which the permanent
$$ \operatorname{permanent}(A) = \sum _{\pi \in S_n} a_{1, \pi(1)} a_{2, \pi(2)} \cdots a_{n, \pi(n)} = 1 $$
Are all the stochastic matrices with permanent equal to $1$ permutation matrices?
Here's a probabilistic proof.
Suppose we have $n$ independent RVs $X_1,\dots,X_n$ each distributed on $\{1,\dots,n\}$ according to the respective row probability vector $a_{1,\cdot}, \dots, a_{n,\cdot}$. Then
$$\mbox{Perm}(A) = P(X_1,\dots, X_n \mbox{ are distinct}).$$
By assumption both sides are equal to $1$.
As a result if $P(X_i= j)=a_{i,j}>0$, then $P(X_{i'}=j)=a_{i',j}=0$ for all $i'\ne i$, and therefore each column has at most one non-zero element. As every row has at least one non-zero element, the result follows.
Yes, they are. Let $A$ be stochastic square matrix as below and $\operatorname{perm}(A) = 1$. Observe that
$$1 = \operatorname{perm}(A) = \sum _{f \in S_n} \prod_{i=1}^n a_{i, f(i)} \leq \sum_{f \colon [n] \to [n]} \prod_{i=1}^n a_{i, f(i)} = \sum _{f_1, \ldots, f_n = 1}^n \prod_{i=1}^n a_{i, f_i} = \prod_{i=1}^n \left(\sum _{j=1}^n a_{k,j}\right) \leq 1 $$
where we fist just added more (positive) elements to the sum, and use the fact that $\sum\limits_{j=1}^n a_{k,j} \leq 1$.
Subtract $1$ of both sides of inequality, and get
$$0 \leq \sum _{f \colon [n] \to [n]\\f \notin S_n} \prod_{i=1}^n a_{i, f(i)} \leq 0$$
For every pair $i \neq j$, and any column index $k$. we can prove that (at least) one of $a_{i,k}$ and $a_{j, k}$ must be zero:
It follows from the inequality above that $0 = \prod_{i=1}^n a_{i, f(i)} $ for any $f$, which is not bijective.
Take such $f$, that $f(i)=f(j)=k$ and $a_{s, f(s)} \neq 0$ for $s \notin \{i, j\}$ (such function exists, because at least one element in each row must be nonzero). That means that one of $a_{i,k}$ or $a_{j,k}$ must be zero. From that it is clear to see that in each column, no more than one element must be nonzero.
We may prove by mathematical induction that the permanent of every substochastic matrix $A$ is at most $1$, and equal to $1$ if and only if $A$ is a permutation matrix.
The base case for $1\times1$ substochastic matrices is trivial. In the inductive step, suppose $A$ is at least $2\times2$. Denote by $M_{ij}$ the submatrix obtained by deleting the $i$-th row and the $j$-th column of $A$. Then each $M_{ij}$ is substochastic and $$ \operatorname{per}(A)=\sum_ja_{ij}\operatorname{per}(M_{ij})\le\sum_ja_{ij}\le1. $$ When $\operatorname{per}(A)=1$, we must have $\sum_ja_{ij}=1$, and $\operatorname{per}(M_{ij})=1$ whenever $a_{ij}>0$. In particular, $A$ is row-stochastic. Now, if $A$ has two positive elements on the $i$-th row, then $\operatorname{per}(M_{ij})=1$ for some two different $j$s, say $j_1$ and $j_2$. Therefore, by induction assumption, both $M_{ij_1}$ and $M_{ij_2}$ are permutation matrices. It follows that $A_{kj_1}=A_{kj_2}=1$ for some $k$, but this is impossible because $A$ is row-stochastic. Hence $A$ has exactly one positive element on each row. As $M_{ij}$ is a permutation matrix whenever $a_{ij}>0$, $A$ is a permutation matrix. Conversely, if $A$ is a permutation matrix, obviously its permanent is $1$.
