Let $N(q,n)=\frac{1}{n}\sum_{d|n}\mu(n/d)q^d$ for $q$ positive integer. Is it true that $N(q,n)<q^n/n$? This is true for $q$ prime which corresponds to the number of monic irreducible polynomials of degree $n$ over field $F_q$.
I feel like this should be easy to see from the sum since $\mu(1)q^n=q^n$ and this is the largest term in the sum but I can't convince myself that the other lower order terms somehow don't overflow the bound. Unfortunately the sum is not multiplicitive. Numerically this looks true up to $n=100$.
I find it easier to switch the roles of $d$ and $n/d$, so I write
$$N(q,n) = \frac{1}{n}\sum_{d\mid n} \mu(d) q^{n/d}.$$
It is immediate that for $n = 1$ we have $N(q,1) = q$, so in that case the strict inequality does not hold. For $n > 1$, let $p$ be the smallest prime factor of $n$. If $n = p$, we have $N(q,p) = \frac{1}{p}(q^p-1) < \frac{1}{p}q^p$, so the strict inequality holds. If $n$ is not prime, we have
\begin{align} n\cdot N(q,n) &= \sum_{d\mid n} \mu(d) q^{n/d}\\ &= q^n - q^{n/p} + \sum_{\substack{d\mid n\\ d > p}} \mu(d)q^{n/d}\\ &\leqslant q^n - q^{n/p} + \sum_{\substack{d\mid n\\ d > p}} q^{n/d}\\ &\leqslant q^n - q^{n/p} + \sum_{k=0}^{\frac{n}{p}-1} q^k\\ &= q^n - q^{n/p} + \frac{q^{n/p}-1}{q-1}\\ &< q^n, \end{align}
as desired.