As far as I understand, you suppose that $\partial A$ is disjoint from $A$, and in the process you define $A_n=A_{n-1}\cup \partial A_{n-1}$,
Under these assumptions, the answer to both questions is negative.
Consider a binary tree with a root vertex $a$ (that is, $a$ has degree 1, and each other vertex has degree 3). Next, for every positive integer $d$, identify all the vertices at distance $3d$ from $a$. a$ (obtaining a vertex $v_d$; the vertices $v_d$ are different for differet $d$). Finally, take two disjoint copies of this graph and identify their root vertices (denote the resulting vertex by $a$ again).
Now, $j=j(a)=0$ since $|\partial A_{3d-1}|=2$. On the other hand, let $b$ be one of the two neighbors of $a$. Then $|A_n|\leq 2^{n+2}$, but $|\partial A_n|\geq 2^{n-1}$, so $j(b)\geq 1/8$. Thus we have constructed a counterexample to both conjectures.
On the other hand, you may ask the same questions for a graph with a bounded degree. Then the answer to the second question is affirmative, though there still exist counterexamples to the first one.
Here is a counterexample in this case. Instead of glueing vertices, we will thin out a tree. First, take the same binary tree with a rot vertex $a$. Now, for some $n_1$ large enough, delete some vertices of the $n_1$th layer so that $|\partial A_{n_1-1}|\approx |A_{n_1-1}|/1000$ (we delete each vertex together with a subtree hanging on it). Note that now we have, say, $|\partial A_{n_1}|\geq |A_{n_1}|/800$. Next, take $n_2$ much larger than $n_1$ and thin out $n_2$th layer in the same manner, and so on. Finally, glue two copies of such tree at vertex $a$ again. The we have $j(a)=1/1000$ but $j(b)>1/1000$ if $b$ is a neighbor of $a$.
