I wrote an answer in the comment section, so I'll rewrite it here.
The claim of
$$F_1(q)-F_3(q)\equiv F_1(q^p) \pmod p$$
is equivalent to the following two cases
$$p\nmid n: \sum_{d|n} d-nd^3 \equiv 0 \pmod p$$
$$p|n: \sum_{d|n} d-nd^3 \equiv \sum_{d| \frac{n}{p}} d \pmod p$$
The second case is trivial, because if $p|n$, we have
$$\begin{split} \sum_{d|n} d-nd^3 & \equiv \sum_{d|n} d \pmod p \\ &\equiv \sum_{d|\frac{n}{p}} d \pmod p\end{split}$$
Thus we only need to check
$$p\nmid n: \sum_{d|n} d-nd^3 \equiv 0 \pmod p$$
For $p=2$, this is immediate as it holds for each term in the sum individually. That is, we necessarily have $n \equiv 1 \pmod 2$, and $d-nd^3 \equiv d-d^3 \equiv d(1-d)(1+d) \equiv 0 \pmod 2$.
For $p=3$ and $p=5$, it is worthwhile to write
$$\sum_{d|n} d-nd^3 = \sum_{d|n\text{ and } d<\sqrt{n}} (d-nd^3)+\left(\frac{n}{d}-n\left(\frac{n}{d}\right)^3\right)$$
My claim is that each term in the sum on the right is congruent to $0$ mod $3$ and $5$.
This follows from
$$(d-nd^3)+\left(\frac{n}{d}-n\left(\frac{n}{d}\right)^3\right) = \frac{1}{d^3}(d^2+n)(d^2-nd^4+n^2d^2-n^3)$$
We can now proceed by cases, remembering that $p \nmid n$ and hence $p \nmid d$.
For $p=3$, first note that $d^2 \equiv 1 \pmod 3$.
Then we enjoy
$$\begin{split} \frac{1}{d^3}(d^2+n)(d^2-nd^4+n^2d^2-n^3) &\equiv d (1+n)(1-n+n^2-n^3) \pmod 3 \\ \equiv d(1+n)(1-n)(1+n^2) \pmod 3 \end{split}$$
Note that we have $n \equiv 1 \pmod 3$ or $n \equiv 2 \pmod 3$, and plugging those in gives congruence to $0$.
Thus we have confirmed the $p=3$ case.
For $p=5$, first note that $d^2 \equiv 1 \pmod 5$ or $d^2 \equiv 4 \pmod 5$.
In the first case,
$$\begin{split} \frac{1}{d^3}(d^2+n)(d^2-nd^4+n^2d^2-n^3) & \equiv d(1+n)(1-n)(1+n^2) \pmod 5 \end{split}$$
In the second case,
$$\begin{split} \frac{1}{d^3}(d^2+n)(d^2-nd^4+n^2d^2-n^3) & \equiv d(4+n)(4-n)(1+n^2) \pmod 5 \end{split}$$
It's straightforward to check that $n \equiv 1,2,3,4 \pmod 5$ plugged into the above will vanish.
Thus we have confirmed the $p=5$ case.
We can rule out the claim of $$F_1(q)-F_3(q)\equiv F_1(q^p) \pmod p$$ for all other $p$ by noting that for $n=2$,
$$\sum_{d|n} d-nd^3 \equiv -15 \pmod p$$
This is a problem for prime $p>5$, as then we have $p \nmid n$ but $\sum_{d|n} d-nd^3$ is not congruent to $0$.
(As an aside, note that this isn't a problem for $p=2$, since then $p|n$ and it's straightforward to check that it's congruent to $\sum_{d|\frac{n}{p}} d$, as needed. This isn't a problem for $p=3,5$, since then $p \nmid n$ and we clearly have congruence to $0$, as needed.)
EDIT: I realized that in proving the $p=3,5$ cases for $p \nmid n$ that there was an edge case I needed to check. I check this edge case here.
To prove the cases of $p=3,5$, I used
$$\sum_{d|n} d-nd^3 = \sum_{d|n\text{ and } d<\sqrt{n}} (d-nd^3)+\left(\frac{n}{d}-n\left(\frac{n}{d}\right)^3\right)$$
This works whenever $\sqrt{n}$ is not an integer. When $\sqrt{n}$ is an integer, it's a valid divisor, and in that case we'll have one more term
$$\sum_{d|n} d-nd^3 = \sqrt{n}-n(\sqrt{n})^3 + \sum_{d|n\text{ and } d<\sqrt{n}} (d-nd^3)+\left(\frac{n}{d}-n\left(\frac{n}{d}\right)^3\right)$$
So far, I showed that the rightmost sum is congruent to $0$ mod $3$ and $5$, so I only need to check this extra term. This extra term is
$$ \sqrt{n}-n(\sqrt{n})^3 = \sqrt{n}(1-n)(1+n)$$
For the case of $p=3$, $n \equiv 1 \pmod 3$, since $n$ is a square number not divisible by $3$. For the case of $p=5$, $n \equiv 1 \pmod 5$ or $n \equiv 4 \pmod 5$, since $n$ is a square number not divisible by $5$. Plugging these values in confirms congruence to $0$.
