Characterize stochastic matrices such that max singular value is less or equal one.

Here is a complete characterisation:

Proposition. Suppose $A$ is row stochastic (hence we always have $1=\rho(A)\le\|A\|_2$). Then $\|A\|_2=1$ if and only if $A$ is doubly stochastic.

Proof. If $A$ is doubly stochastic, so is $A^TA$ and hence $\left(\|A\|_2^2\equiv\right)\rho(A^TA)=1$.

Conversely, if $\|A\|_2(\equiv\|A^T\|_2)=1$, let $e$ be the all-one vector. Then $$ \|e\|_2^2=\langle e,e\rangle=\langle e,Ae\rangle=\langle A^Te,\,e\rangle\le\|A^Te\|_2\|e\|_2\le\|A^T\|_2\|e\|_2^2=\|e\|_2^2. $$ Therefore ties must occur in the two inequalities above, meaning that $A^Te$ is parallel to $e$ and $\|A^Te\|_2=\|e\|_2$. Thus $A^Te=e$, i.e. $A$ is also column-stochastic.

Perron Frobenius theory says the largest eigenvalues is 1. I believe at this point, you just need the matrix to be normal, so that the spectral radius equals the operator norm.

Since the singular values of $A$ are the square roots of the positive eigenvalues of $A^tA$ then a necessary and sufficient is the eingevalues of $A^tA$ smaller or equal to 1. However, I believe this is not the answer that you want. So let me give you a sufficient condition which in some sense is sharp.

A sufficient condition for the singular values of a stochastic matrix $A\in M_n(\mathbb{R})$ to be smaller or equal to 1 is the columns sums smaller or equal to 1.

Remind that the singular values of $A$ are the square roots of the positive eigenvalues of $A^tA$. Notice that $A^tA$ is also a non negative matrix and the biggest eigenvalue of $A^tA$ is smaller or equal to the biggest row sum (See item 8 here in the subsection positive matrices).

Now notice that the row sums of $A^tA$ are the entries of the vector $A^tAu$, where $u=(1,\ldots,1)\in\mathbb{R}^n$. But $A^tAu=A^tu$, since $A$ is stochastic. Thus, the row sums of $A^tA$ are the row sums of $A^t$, which are the column sums of $A$.

Now, notice that there are examples of stochastic matrices whose columns sums are smaller or equal to $1+\epsilon$ and the singular values are bigger than 1. Thus, in some sense the sufficient condition above is sharp. For example: $uv^t$, where $v=(\frac{1+\epsilon}{n},\frac{1-\epsilon}{n},\frac{1}{n},\ldots,\frac{1}{n})\in\mathbb{R}^n$. Notice that $|v|=\frac{\sqrt{n+2\epsilon^2}}{n}$, $|u|=\sqrt{n}$ and the singular value of $uv^t$ is $|u||v|=\frac{\sqrt{n+2\epsilon^2}}{\sqrt{n}}>1$.