Define
$$a(n) = \frac{1}{n} \sum_{ d|n } \mu{\left(\frac{n}{d}\right)} 2^d $$
where $\mu()$ is the Möbius function.
Is it possible to find easily computable $b, c$ such that $b(n) \leq a(n) \leq c(n)$ for all large enough $n$ such that
$$\limsup \frac{a(n)}{c(n)} = 1 = \liminf \frac{a(n)}{b(n)}.$$
The first few values of $a(n)$ are 1, 2, 1, 2, 3, 6, 9, 18, 30, 56, 99, 186, 335, 630, 1161, 2182, 4080, 7710, 14532, 27594, 52377, 99858.
For $d = n$, we always have the contribution
$$\mu\left(\frac{n}{n}\right)2^n = 2^n$$
to the sum. The remaining sum satisfies
$$\left\lvert \sum_{\substack{d\mid n\\ d < n}}\mu\left(\frac{n}{d}\right)2^d\right\rvert\leqslant \sum_{\substack{d\mid n\\ d < n}} 2^d \leqslant \tau(n)2^{n/2} \leqslant 2\sqrt{n}2^{n/2}$$
since $d\leqslant \frac{n}{2}$ for all divisors $< n$, and there are no more that $2\sqrt{n}$ divisors of $n$. (We can take a better bound on $\tau(n)$ to obtain tighter bounds, but off-hand, I don't recall the best bounds on $\tau$.)
Then we can take
$$b(n) = \frac{1}{n}\left(2^n - 2\sqrt{n}2^{n/2}\right) \leqslant a(n) \leqslant \frac{1}{n}\left(2^n + 2\sqrt{n}2^{n/2}\right) = c(n),$$
and since $b\sim c$ have the asymptotic equality
$$a(n) \sim \frac{2^n}{n}.$$