If ${q^k}\sigma(q^k) \equiv a \pmod b$ and $\displaystyle\frac{\sigma(q^k)}{n} \neq \displaystyle\frac{\sigma(n)}{q^k}$, does it follow that $n\sigma(n) \not\equiv a \pmod b$?
Here, $q$ is prime and $n$ is composite.
If ${q^k}\sigma(q^k) \equiv a \pmod b$ and $\displaystyle\frac{\sigma(q^k)}{n} \neq \displaystyle\frac{\sigma(n)}{q^k}$, does it follow that $n\sigma(n) \not\equiv a \pmod b$?
Here, $q$ is prime and $n$ is composite.
We can generate counterexamples at will. Take for example $n=4$. Then $n\sigma(n)=28$.
Take $q=11$, $k=1$. Then $q^k\sigma(q^k)=132$. We find $a$ and $b$ such that $28\equiv a \pmod{b}$ and $132\equiv a\pmod{b}$. Choose $a=2$. Then $28\equiv 2\pmod{13}$ and $132\equiv 2\pmod{13}$. It is too easy to arrange for two numbers to be each congruent to $a$ modulo $b$ if we are allowed to choose $a$ and $b$. A simpler class of counterexamples would use $b=1$.
Added: For an example with $a=2$, $b=4$ as asked for in a comment, let $q=17$, $k=1$, and $n=15$. Then $q^k\sigma(q^k)\equiv 2\pmod{4}$ but $n\sigma(n)\not\equiv 2\pmod{4}$.