(I)Solution:We first show that $f(A)\leq A$,the equation holds iff $A=1,2,3,4,5,6,7,8,9,19$. $A=(a_na_{n-1}\ldots a_0),(a_n\neq0)$. If $n\geq2$,$A\geq10^n\geq25\times2^n$,$f(A)\leq9(1+2+\ldots+2^n)=9(2^{n+1}-1)<18\times2^n<25 \times2^n$.$\therefore f(A)<A$. If $n=0$,it's clear that $f(A)=A$.$A=1,2,3,4,5,6,7,8,9$. If $n=1$,$A=10a_1+a_0,f(A)=2a_0+a_1,(0\leq a_0\leq9,1\leq a_1\leq9)$. Then $f(A)-A=a_0-9a_1\leq a_0-9\leq0$.The equation holds if and only if $a_0=9,a_1=1$. So $f(A)\leq A$,the equation holds iff $A=1,2,3,4,5,6,7,8,9,19$. If there doesn't exist k such that $A_{k+1}=A_k$,then since $A_{k+1}=f(A_k)\leq A_k$,$A_{k+1}<A_k$. So $A_{A+1}\leq A-A=0$,a contradiction. So there must exist k,$A_{k+1}=A_k$. (II)For A=19^{86},it's easy to know that $A_k\in\{1,2,3,4,5,6,7,8,9,19\}$. We show that if 19|A,then 19|f(A). $A=(a_na_{n-1}\ldots a_0),(a_n\neq0)$. \[ A=a_n10^n+a_{n-1}10^{n-1}+\ldots+a_0=\sum_{i=0}^na_i10^i \] \[ f(A)=\sum_{i=0}^n2^{n-i}a_i \] So \[ 2^nA-f(A)=\sum_{i=0}^na_i20^i\times2^{n-i}-\sum_{i=0}^n2^{n-i}a_i=\sum_{i=0}^na_i2^{n-i}(20^i-1) \] And for $i=0,1,\ldots,n$,$19|20^i-1$,so $19|2^nA-f(A)$,and $\gcd(19,2^n)=1$,so if $19|A$,$19|f(A)$. And $19|A=19^{86}$,so $19|A_k$,and $A_k=19$.