**Edit.** The approach I originally suggested, that is, leveraging Lemma 1 below to carry a case analysis, isn't as promising as I thought. In this updated "answer" I provide a detailed proof of the following proposition:
> **Proposition.** Let $k, n \ge 1$ and let $A_1, \dots, A_k$ be $k$ multisets of size $n$ with all their elements in the interval $[0, 1]$.
Then we can turn $A_1, \dots, A_k$ into $k$ multisets $A_1'\, \dots, A_k'$ of size $n$ such that $\vert \sigma(A_i') - \sigma(A_j') \vert \le 1$ for every
$i, j\in \{1, \dots, k\}$ by means of at most $\lfloor \frac{n}{2} \rfloor + \lfloor \frac{2n}{3} \rfloor + \cdots + \lfloor \frac{(k - 1)n}{k} \rfloor$ swaps.
Denoting by $m(k, n)$ the minimum number of swaps required in the worst case, the above proposition yields $$m(4, n) \le \frac{23}{12}n.$$
Our proof follows the ordering based approach suggested by the OP at the bottom of his post.
> *Proof.* We proceed by induction on $k \ge 1$. If $k = 1$, the result is obvious. Let us assume that $k > 1$ and
let $(a_l)_{1 \le l \le kn}$ be the sequence of real numbers which consists of the elements of $\bigsqcup_{i = 1}^k A_i$ sorted in increasing order.
For $i \in \{1, \dots, k\}$, we define $A_i' = \{ a_l \,\vert\, l \equiv i \mod k \}$ which is a multiset of size $n$. For every $i, j \in \{1, \dots, k\}$ such that $i < j$, we observe that
$\sigma(A_j') - a_{(k - 1)n + j} \le \sigma(A_i') \le \sigma(A_j')$ so that $\vert \sigma(A_i') - \sigma(A_j') \vert \le 1$ for every
$i, j\in \{1, \dots, k\}$. As $\bigsqcup_{i = 1}^k A_i = \bigsqcup_{i = 1}^k A_i'$, there is at least one integer $i \in \{1, \dots, k\}$ such that
$\vert A_i \cap A_1' \vert \ge \frac{n}{k}$. We can assume, without loss of generality that $i = 1$.
After turning $A_1$ in to $A_1'$ by means of at most $\lfloor \frac{(k - 1)n}{k} \rfloor$, we can apply the induction hypothesis to the multisets $A_2, \dots, A_k$, which completes the proof.
In the above proof, we have reduced OP's problem to the following:
> **Problem.** Let $k$ sets with $n$ balls be such that balls in the same set have the same colour and distinct sets have distinct colours. Fill $k$ bins of capacity $n$ with the $kn$ coloured balls.
In the worst case, what is the smallest number $\mu(k, n)$ of swaps that render all bins monochrome?
Thanks to the problem reduction, we have $$m(k, n) \le \mu(k, n)$$ and the proposition yields $$\mu(k, n) \le (k - H(k))n$$ where $H(k) := 1 + \frac{1}{2} + \cdots + \frac{1}{k}$.
> **Claim.** We have $\mu(2, n) = \lfloor \frac{n}{2} \rfloor$ and $n - 2 \le \mu(3, n) \le n$.
> *Poof.* Proving the identity $\mu(2, n) = \lfloor \frac{n}{2} \rfloor$ is straightforward. Let us show that $\mu(3, n) \le n$. Let $r_i, g_i$ and $b_i$ be respectively the number of red, green and blue balls in the $i$-th bin for $i \in \{1, 2, 3\}$. Relabelling the bins and permuting the colours if needed, we can assume that $r_1 = \max \{r_i, g_i, b_i \, \vert \, i =1, 2, 3\}$. Since $g_1 + b_1 = r_2 + r_3$, we may assume, without loss of generality that (1) $g_1 \le r_2$ or (2) $b_1 \le r_2$ (simply observe that $\min(g_1, b_1) \le \max(r_2, r_3)$).
We shall address the case (1) only, as (2) is similar.
We perform $g_1 + b_1$ swaps to turn the first bin into a red monochrome in such a way that the $g_1$ green balls of the first bin go into the second bin. Now the second bin contains $g_1 + g_2$ green balls so that it takes $g_3$ swaps to render it green monochrome. Since $g_3 \le r_1$, the case (1) is settled. To show that $\mu_(3, n) \ge n - 2$, it suffices to consider the filling for which the distribution of colours in each bin is the closest from the uniform distribution.
> **Corollary.** We have $\frac{2}{3}n \le m(3, n) \le n$.
---
*The remains of my original answer.*
This is a **long comment** hinting towards a case analysis.
Given a finite multiset $A$ consisting of real numbers, we denote by $\sigma(A)$ the sum of its elements.
We settle below the case of **two multisets**.
The details (which are likely to be well-known) are then used to provide a non-trivial upper bound in the case of **three multisets**.
They could also be used to conduct a case analysis in the situation of four multisets.
> **Lemma 1.** Let $A$ and $B$ be two multisets of size $n$ with all their elements in $[0, 1]$ and such that
$\sigma(A) \ge \sigma(B)$. Let $\lambda \in [0, 1]$.
Then we can turn $A$ into a multiset $A'$ of size $n$ such that
$$\vert \sigma(A') - ((1 - \lambda) \sigma(A) + \lambda \sigma(B)) \vert \le \frac{1}{2}$$
by swapping at most $\lceil \lambda n\rceil$ elements of $A$ with some elements of $B$.
> *Proof.* Sort $A$ in decreasing order and let $M_i$ be the $i$-th term of the sorted sequence.
Sort $B$ in increasing order and let $m_i$ be the $i$-th term of the sorted sequence.
Let $s_0 = \sigma(A)$, and for $i \ge 1$, let $s_i$ be the sum of the elements of the multiset $A'_i$ obtained from $A$
by swapping $M_1, \dots, M_i$ with $m_1, \dots, m_i$ respectively.
The $i$-th second order [finite difference](https://en.wikipedia.org/wiki/Finite_difference) $\Delta_2(s_i)$ of $(s_j)_{0 \le j \le n}$ is $$m_{i + 1} - m_i - (M_{i + 1} - M_i),$$
which is non-negative for every $i \in \{0, \dots, n - 2\}$.
Therefore the piecewise linear interpolation $s$ of $(s_j)_{0 \le j \le n}$ on $[0, n]$ is a convex function.
As $s$ satisfies $s(0) = \sigma(A)$ and $s(n) = \sigma(B)$, the convexity of $s$ implies that
$s(\lambda n) \le \mu$ where $\mu := (1 - \lambda) \sigma(A) + \lambda \sigma(B)$. To complete the proof, we consider
$x_{\mu} := \inf \{ x \in [0, 1] \,\vert\, s(x) > \mu\}$. Since $\vert s_{i + 1} - s_i\vert \le 1$ for every $i \in \{0, \dots, n -1\}$,
at least one of $A_{\lfloor{x_{\mu}} \rfloor}$ and $A_{\lceil{x_{\mu}} \rceil}$ is at distance at most $\frac{1}{2}$ of $\mu$.
> **Lemma 2.** Let $a, b$ and $c$ be non-zero real numbers such that $a + b + c = 0$. There is a permutation $\theta$ of the letters $\{a, b, c\}$
such that $\left\vert \frac{\theta(a)}{ \theta(b)} \right\vert \ge 2.$
> *Proof.* Renaming the variables if needed, can assume that $b$ and $c$ have the same sign and that $\vert b \vert \le \vert c \vert$.
The result now follows from the identity $\left\vert\frac{a} {b} \right\vert = 1 + \left\vert\frac{c} {b} \right\vert$.
> **Claim (Not verified, hence possibly wrong).** Let $A, B$ and $C$ be three multisets of size $n$ with all their elements in the interval $[0, 1]$.
We can turn these multisets into three multisets $A', B'$ and $C'$ of size $n$ and such that $$\vert \sigma(M) - \sigma(N) \vert \le 1 \text{ for every} M, N \in \{A', B', C'\}$$
by means of at most $\lceil \frac{n}{2} \rceil + \lceil \frac{n}{3} \rceil$ swaps between multiset elements.
**Note.** The proof below is **incorrect**. Indeed, it actually enforces only two of the three required conditions. For the third condition, we get only an approximation up to $\frac{5}{4}$. It is possible to fix the proof, but at the cost of increasing the number of required swaps in a very disappointing way: $2\lceil \frac{n}{2} \rceil + \lceil \frac{n}{3} \rceil$.
> *Proof.* Let $S = \sigma(A) + \sigma(B) + \sigma(C)$.
Let us assume first that $\sigma(A) = \frac{S}{3}$. Renaming the multisets if needed, we can assume that $\sigma(C) \le \frac{S}{3} \le \sigma(B)$.
Since obviously $\frac{\sigma(B) + \sigma(C)}{2} = \frac{S}{3}$, we can use Lemma 1 with $\lambda = \frac{1}{2}$ to enforce all the conditions of the claim
after at most $\lceil \frac{n}{2} \rceil$ swaps between elements of $B$ and $C$.
>We can, and shall assume from now on, that none of $\sigma(A), \sigma(B)$ and $\sigma(C)$ is equal to $\frac{S}{3}$.
Renaming the multisets if needed, we can assume that, either
(1) $\sigma(C) < \frac{S}{3} < \sigma(B) \le \sigma(A)$
or
(2) $\sigma(C) \le \sigma(B) < \frac{S}{3} < \sigma(A)$.
>Let us address (1) first. By Lemma 2 and its proof, we have $\frac{\frac{S}{3} - \sigma(C)}{\sigma(B) - \frac{S}{3}} \ge 2$
so that $\lambda := \frac{\sigma(B) - \frac{S}{3}}{\sigma(B) - \sigma(C)} \le \frac{1}{3}$.
We apply Lemma 1 to $B, C$ and the previous value of $\lambda$, turning $B$ into a multiset $B'$ such that $\vert \sigma(B') - \frac{S}{3} \vert \le \frac{1}{2}$
by means of at most $\lceil \frac{n}{3} \rceil$ swaps between the elements of $B$ and $C$. After applying those swaps, the multiset $C$ has been turned into a multiset $C'$ satisfying $\left\vert \frac{\sigma(A) + \sigma(C')}{2} - \frac{S}{3} \right\vert \le \frac{1}{4}$.
We conclude this case by applying Lemma 1 to $A$ and $C'$ with $\lambda = \frac{1}{2}$.
Let us now address (2). By Lemma 2 and its proof, we have $\frac{\sigma(A) - \frac{S}{3}}{\frac{S}{3} - \sigma(B)} \ge 2$ so that
$\lambda : = \frac{\frac{S}{3} - \sigma(B)}{\sigma(A) - \sigma(B)} \le \frac{1}{3}$. We apply Lemma 1 to $A, B$ and the previous value of $\lambda$, turning
$B$ into a multiset $B'$ such that $\vert \sigma(B') - \frac{S}{3} \vert \le \frac{1}{2}$ by means of at most $\lceil \frac{n}{3} \rceil$
swaps between the elements of $A$ and $B$. After applying those swaps, the multiset $A$ has been turned into a multiset $A'$ satisfying
$\left\vert \frac{\sigma(A') + \sigma(C)}{2} - \frac{S}{3} \right\vert \le \frac{1}{4}$.
We conclude this case by means of an application of Lemma 1 to $A'$ and $C$ with $\lambda = \frac{1}{2}$.