I discovered that my conjecture is incorrect. Here is a simple counterexample.

First, let's allow $A_{ij}=0$ for some $i,j.$ Let

$$A_{ij} = \begin{cases}\frac{1-\epsilon}{(k-1)^2} & 1\leq i,j\leq k-1 \\ \epsilon & i=j=k \\ 0 & \mbox{else}. \end{cases}$$

Then, $\|B\| = 1$ since $B_{kk} = 1,$ and we know that $\|B\|\in[1/k,1].$ However, $\|B^\prime\| = \frac{1}{k-1}.$

To get a counterexample with all $A_{ij}$ strictly positive, let

$$A_{ij} = \begin{cases}\frac{1-\epsilon}{(k-1)^2}-\delta & 1\leq i,j\leq k-1 \\ \epsilon-(k-1)\delta & i=j=k \\ \delta & \mbox{else} \end{cases},$$

and choose $\delta$ much much smaller than $\frac{\epsilon}{k}.$

I discovered that my conjecture is incorrect. Here is a simple counterexample.

First, let's allow $A_{ij}=0$ for some $i,j.$ Let

$$A_{ij} = \begin{cases}\frac{1-\epsilon}{(k-1)^2} & 1\leq i,j\leq k-1 \\ \epsilon & i=j=k \\ 0 & \mbox{else}. \end{cases}$$

Then, $\|B\| = 1$ since $B_{kk} = 1,$ and we know that $\|B\|\in[1/k,1].$ However, $\|B^\prime\| = \frac{1}{k-1}.$

To get a counterexample with all $A_{ij}$ strictly positive, let

$$A_{ij} = \begin{cases}\frac{1-\epsilon}{(k-1)^2} & 1\leq i,j\leq k-1 \\ \epsilon-(2k-2)\delta & i=j=k \\ \delta & \mbox{else} \end{cases},$$

and choose $\delta$ much much smaller than $\frac{\epsilon}{k}.$