Let $\chi$ be a Dirichlet character, and define $\phi_\chi (n)$ so that it satisfies $$\sum_{n=1}^\infty \phi_\chi (n)n^{-s}=\frac{\zeta(s-1)}{L(s,\chi)}.$$

In other words

$$\phi_{\chi}(n)=\sum_{d|n}\mu\left(\frac{n}{d}\right)\chi\left(\frac{n}{d}\right)d=\left(\text{Id}*\mu\chi\right)(n).$$

My question is, how large can $\phi_\chi(n)$ be? More precisely, what is the smallest function $f$ such that for all $n$ and $\chi$ $$\frac{\phi_{\chi}(n)}{n}=\sum_{d|n}\frac{\mu(d)\chi(d)}{d}\ll f(n).$$

It is not hard to see that $\frac{\phi_{\chi}(n)}{n}\ll \log n$ for all $n$ so $f\ll \log n$. For Euler's $\phi$ function, the first term is the main contributor, as $\phi(n)\leq n$ we know that $\frac{\phi(n)}{n}\leq 1$ for all $n$. Since $\chi$ has norm $1$, and the sums $\sum_{n\leq x}\mu(n)\chi(n)$ are small, we might conjecture that $\frac{\phi_\chi (n)}{n}\leq 1$.

**However this is not so.** We can find a character such that $\mu$ and $\chi$ function have a lot of correlation, enough to make the sum of size $\sqrt{\log\log n}$. This is outlined by the following construction:

Let $n$ be the product of all primes $p=3+4k$ where $p\leq M$, and let $\chi$ be the unique quadratic character modulo $4$. Then choose whether or not to remove the prime $3$ from this product as to force the equivalence $n\equiv 1 \pmod{4}$. For each divisor $d$ of $n$, if $\omega(d)$ is even, then $d\equiv 1\pmod{4}$ so that $\chi(d)\mu(d)=1$, and if $\omega(d)$ is odd, then $d\equiv 3\pmod{4}$ so that $\chi(d)\mu(d)=1$ yet again. This means that $$\frac{\phi_\chi (n)}{n}=\sum_{d|n} \frac{1}{d}\gg \sqrt{\log M}\gg\sqrt{\log \log n}.$$ The second last $\gg$ follows from the fact that if $A$ is the set of integers composed only of primes congruent to $3$ modulo $4$, then $\sum_{n\leq M,\ n\in A} \frac{1}{n}=\sqrt{\log M}$, and the last $\gg$ follows from the fact that $\log n =\theta(M;4,3)$.

Any references to papers which might deal with this sort of sum is greatly appreciated,

Thanks,