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UPDATE: As pointed out by James, this attempt is not convincing. The procedure I suggested in the last paragraph is not really well defined. I still believe that $\sim$ is transitive, and that a more careful ``common refinement'' type argument should work. For now, I can only show that James's example is not a counterexample.

We have $$ \mu = \sum_{n=0}^{\infty} 2\delta_{2n+1}, \quad \nu = \delta_0 + \delta_1 + \sum_{n=1}^{\infty} 2\delta_{2n+1} . $$ Both measures are equivalent to a common third measure, so we want to show that $\mu\sim\nu$. This we can do by working our way from the left to the right.

First of all, split off $\delta_0+(1/2)\delta_3$ from $\nu$, and pair this with $(3/2)\delta_1$, and also split off $(1/2)\delta_1$ from each measure. Now the remaining part of $\nu$ starts with $(1/2)\delta_1 + (3/2)\delta_3 + 2(\delta_5+\delta_7+\ldots)$, while $\mu$ has been reduced to $2(\delta_3+\delta_5+\ldots)$.

Next, pair $(1/2)(\delta_1+\delta_5)$ (from $\nu$) with $\delta_3$, and also split off $\delta_3$ from both measures. I'm now in the same configuration as before this step, but everything has been shifted to the right by two units. So I can continue indefinitely in this style.

ORIGINAL (IMPRECISE) ANSWER STARTS HERE: The relation $\sim$ is transitive. The argument is elementary, but rather tedious to spell out precisely, so I'll be light on the details.

Observe first of all that by decomposing further if necessary, we can get to a situation where all comparisons $\mu_j\leftrightarrow\nu_j$ are between two atoms $\mu_j=\nu_j=g\delta_n$ or between an atom and a measure supported by two points. This follows because I can decrease the number of points in the combined supports by decomposing further if I'm not in this situation yet. For example, if $$ a:=\max\textrm{supp}\,\mu_j>\max\textrm{supp}\,\nu_j=:b $$ and (let's say) $\nu_j(b)\le \mu_j(a)$, then I can represent $\nu_j(b)\delta_b$ as a convex combination of part $\mu_j$, and I have eliminated one point from the union of the supports. Similar arguments work in the other cases.

So if $\mu\sim\nu$, I can get from $\mu$ to $\nu$ by splitting and/or combining each point mass $\mu(n)\delta_n$, and for each such operation only two other points are involved. Also, since there are only finitely many other points within reach $A$, finitely many such operations suffice for a given $n$. In other words, I can partition each point mass $\mu(n)=w_1 + \ldots + w_{N_n}$ into finitely many parts, and each part $w_j$ gets a splitting or combining treatment.

Now if $\mu\sim\nu$, $\mu\sim\nu'$, then we can consider the common refinements of these partitions. We obtain corresponding finer decompositions of $\nu$, $\nu'$ (compared to the original ones), and these witness that $\nu\sim\nu'$.

UPDATE: As pointed out by James, this attempt is not convincing. The procedure I suggested in the last paragraph is not really clearly defined. I still believe that $\sim$ is transitive, and that a more careful ``common refinement'' type argument should work. For now, I can only show that James's example is not a counterexample.

We have $$ \mu = \sum_{n=0}^{\infty} 2\delta_{2n+1}, \quad \nu = \delta_0 + \delta_1 + \sum_{n=1}^{\infty} 2\delta_{2n+1} . $$ Both measures are equivalent to a common third measure, so we want to show that $\mu\sim\nu$. This we can do by working our way from the left to the right.

First of all, split off $\delta_0+(1/2)\delta_3$ from $\nu$, and pair this with $(3/2)\delta_1$, and also split off $(1/2)\delta_1$ from each measure. Now the remaining part of $\nu$ starts with $(1/2)\delta_1 + (3/2)\delta_3 + 2(\delta_5+\delta_7+\ldots)$, while $\mu$ has been reduced to $2(\delta_3+\delta_5+\ldots)$.

Next, pair $(1/2)(\delta_1+\delta_5)$ (from $\nu$) with $\delta_3$, and also split off $\delta_3$ from both measures. I'm now in the same configuration as before this step, but everything has been shifted to the right by two units. So I can continue indefinitely in this style.

ORIGINAL (IMPRECISE) ANSWER STARTS HERE: The relation $\sim$ is transitive. The argument is elementary, but rather tedious to spell out precisely, so I'll be light on the details.

Observe first of all that by decomposing further if necessary, we can get to a situation where all comparisons $\mu_j\leftrightarrow\nu_j$ are between two atoms $\mu_j=\nu_j=g\delta_n$ or between an atom and a measure supported by two points. This follows because I can decrease the number of points in the combined supports by decomposing further if I'm not in this situation yet. For example, if $$ a:=\max\textrm{supp}\,\mu_j>\max\textrm{supp}\,\nu_j=:b $$ and (let's say) $\nu_j(b)\le \mu_j(a)$, then I can represent $\nu_j(b)\delta_b$ as a convex combination of part $\mu_j$, and I have eliminated one point from the union of the supports. Similar arguments work in the other cases.

So if $\mu\sim\nu$, I can get from $\mu$ to $\nu$ by splitting and/or combining each point mass $\mu(n)\delta_n$, and for each such operation only two other points are involved. Also, since there are only finitely many other points within reach $A$, finitely many such operations suffice for a given $n$. In other words, I can partition each point mass $\mu(n)=w_1 + \ldots + w_{N_n}$ into finitely many parts, and each part $w_j$ gets a splitting or combining treatment.

Now if $\mu\sim\nu$, $\mu\sim\nu'$, then we can consider the common refinements of these partitions. We obtain corresponding finer decompositions of $\nu$, $\nu'$ (compared to the original ones), and these witness that $\nu\sim\nu'$.

UPDATE: As pointed out by James, this attempt is not convincing. The procedure I suggested in the last paragraph is too simple-minded. I still believe that $\sim$ is transitive, and that a more careful ``common refinement'' type argument should work. For now, I can only show that James's example is not a counterexample.

We have $$ \mu = \sum_{n=0}^{\infty} 2\delta_{2n+1}, \quad \nu = \delta_0 + \delta_1 + \sum_{n=1}^{\infty} 2\delta_{2n+1} . $$ Both measures are equivalent to a common third measure, so we want to show that $\mu\sim\nu$. This we can do by working our way from the left to the right.

First of all, split off $\delta_0+(1/2)\delta_3$ from $\nu$, and pair this with $(3/2)\delta_1$, and also split off $(1/2)\delta_1$ from each measure. Now the remaining part of $\nu$ starts with $(1/2)\delta_1 + (3/2)\delta_3 + 2(\delta_5+\delta_7+\ldots)$, while $\mu$ has been reduced to $2(\delta_3+\delta_5+\ldots)$.

Next, pair $(1/2)(\delta_1+\delta_5)$ (from $\nu$) with $\delta_3$, and also split off $\delta_3$ from both measures. I'm now in the same configuration as before this step, but everything has been shifted to the right by two units. So I can continue indefinitely in this style.

ORIGINAL (FLAWED) ANSWER STARTS HERE: The relation $\sim$ is transitive. The argument is elementary, but rather tedious to spell out precisely, so I'll be light on the details.

Observe first of all that by decomposing further if necessary, we can get to a situation where all comparisons $\mu_j\leftrightarrow\nu_j$ are between two atoms $\mu_j=\nu_j=g\delta_n$ or between an atom and a measure supported by two points. This follows because I can decrease the number of points in the combined supports by decomposing further if I'm not in this situation yet. For example, if $$ a:=\max\textrm{supp}\,\mu_j>\max\textrm{supp}\,\nu_j=:b $$ and (let's say) $\nu_j(b)\le \mu_j(a)$, then I can represent $\nu_j(b)\delta_b$ as a convex combination of part $\mu_j$, and I have eliminated one point from the union of the supports. Similar arguments work in the other cases.

So if $\mu\sim\nu$, I can get from $\mu$ to $\nu$ by splitting and/or combining each point mass $\mu(n)\delta_n$, and for each such operation only two other points are involved. Also, since there are only finitely many other points within reach $A$, finitely many such operations suffice for a given $n$. In other words, I can partition each point mass $\mu(n)=w_1 + \ldots + w_{N_n}$ into finitely many parts, and each part $w_j$ gets a splitting or combining treatment.

Now if $\mu\sim\nu$, $\mu\sim\nu'$, then we can consider the common refinements of these partitions. We obtain corresponding finer decompositions of $\nu$, $\nu'$ (compared to the original ones), and these witness that $\nu\sim\nu'$.

UPDATE: As pointed out by James, this attempt is not convincing. The procedure I suggested in the last paragraph is not really well defined. I still believe that $\sim$ is transitive, and that a more careful ``common refinement'' type argument should work. For now, I can only show that James's example is not a counterexample.

We have $$ \mu = \sum_{n=0}^{\infty} 2\delta_{2n+1}, \quad \nu = \delta_0 + \delta_1 + \sum_{n=1}^{\infty} 2\delta_{2n+1} . $$ Both measures are equivalent to a common third measure, so we want to show that $\mu\sim\nu$. This we can do by working our way from the left to the right.

First of all, split off $\delta_0+(1/2)\delta_3$ from $\nu$, and pair this with $(3/2)\delta_1$, and also split off $(1/2)\delta_1$ from each measure. Now the remaining part of $\nu$ starts with $(1/2)\delta_1 + (3/2)\delta_3 + 2(\delta_5+\delta_7+\ldots)$, while $\mu$ has been reduced to $2(\delta_3+\delta_5+\ldots)$.

Next, pair $(1/2)(\delta_1+\delta_5)$ (from $\nu$) with $\delta_3$, and also split off $\delta_3$ from both measures. I'm now in the same configuration as before this step, but everything has been shifted to the right by two units. So I can continue indefinitely in this style.

ORIGINAL (IMPRECISE) ANSWER STARTS HERE: The relation $\sim$ is transitive. The argument is elementary, but rather tedious to spell out precisely, so I'll be light on the details.

Observe first of all that by decomposing further if necessary, we can get to a situation where all comparisons $\mu_j\leftrightarrow\nu_j$ are between two atoms $\mu_j=\nu_j=g\delta_n$ or between an atom and a measure supported by two points. This follows because I can decrease the number of points in the combined supports by decomposing further if I'm not in this situation yet. For example, if $$ a:=\max\textrm{supp}\,\mu_j>\max\textrm{supp}\,\nu_j=:b $$ and (let's say) $\nu_j(b)\le \mu_j(a)$, then I can represent $\nu_j(b)\delta_b$ as a convex combination of part $\mu_j$, and I have eliminated one point from the union of the supports. Similar arguments work in the other cases.

So if $\mu\sim\nu$, I can get from $\mu$ to $\nu$ by splitting and/or combining each point mass $\mu(n)\delta_n$, and for each such operation only two other points are involved. Also, since there are only finitely many other points within reach $A$, finitely many such operations suffice for a given $n$. In other words, I can partition each point mass $\mu(n)=w_1 + \ldots + w_{N_n}$ into finitely many parts, and each part $w_j$ gets a splitting or combining treatment.

Now if $\mu\sim\nu$, $\mu\sim\nu'$, then we can consider the common refinements of these partitions. We obtain corresponding finer decompositions of $\nu$, $\nu'$ (compared to the original ones), and these witness that $\nu\sim\nu'$.

The relation $\sim$ is transitive. The argument is elementary, but rather tedious to spell out precisely, so I'll be light on the details.

Observe first of all that by decomposing further if necessary, we can get to a situation where all comparisons $\mu_j\leftrightarrow\nu_j$ are between two atoms $\mu_j=\nu_j=g\delta_n$ or between an atom and a measure supported by two points. This follows because I can decrease the number of points in the combined supports by decomposing further if I'm not in this situation yet. For example, if $$ a:=\max\textrm{supp}\,\mu_j>\max\textrm{supp}\,\nu_j=:b $$ and (let's say) $\nu_j(b)\le \mu_j(a)$, then I can represent $\nu_j(b)\delta_b$ as a convex combination of part $\mu_j$, and I have eliminated one point from the union of the supports. Similar arguments work in the other cases.

So if $\mu\sim\nu$, I can get from $\mu$ to $\nu$ by splitting and/or combining each point mass $\mu(n)\delta_n$, and for each such operation only two other points are involved. Also, since there are only finitely many other points within reach $A$, finitely many such operations suffice for a given $n$. In other words, I can partition each point mass $\mu(n)=w_1 + \ldots + w_{N_n}$ into finitely many parts, and each part $w_j$ gets a splitting or combining treatment.

Now if $\mu\sim\nu$, $\mu\sim\nu'$, then we can consider the common refinements of these partitions. We obtain corresponding finer decompositions of $\nu$, $\nu'$ (compared to the original ones), and these witness that $\nu\sim\nu'$.

UPDATE: As pointed out by James, this attempt is not convincing. The procedure I suggested in the last paragraph is too simple-minded. I still believe that $\sim$ is transitive, and that a more careful ``common refinement'' type argument should work. For now, I can only show that James's example is not a counterexample.

We have $$ \mu = \sum_{n=0}^{\infty} 2\delta_{2n+1}, \quad \nu = \delta_0 + \delta_1 + \sum_{n=1}^{\infty} 2\delta_{2n+1} . $$ Both measures are equivalent to a common third measure, so we want to show that $\mu\sim\nu$. This we can do by working our way from the left to the right.

First of all, split off $\delta_0+(1/2)\delta_3$ from $\nu$, and pair this with $(3/2)\delta_1$, and also split off $(1/2)\delta_1$ from each measure. Now the remaining part of $\nu$ starts with $(1/2)\delta_1 + (3/2)\delta_3 + 2(\delta_5+\delta_7+\ldots)$, while $\mu$ has been reduced to $2(\delta_3+\delta_5+\ldots)$.

Next, pair $(1/2)(\delta_1+\delta_5)$ (from $\nu$) with $\delta_3$, and also split off $\delta_3$ from both measures. I'm now in the same configuration as before this step, but everything has been shifted to the right by two units. So I can continue indefinitely in this style.

ORIGINAL (FLAWED) ANSWER STARTS HERE: The relation $\sim$ is transitive. The argument is elementary, but rather tedious to spell out precisely, so I'll be light on the details.

Observe first of all that by decomposing further if necessary, we can get to a situation where all comparisons $\mu_j\leftrightarrow\nu_j$ are between two atoms $\mu_j=\nu_j=g\delta_n$ or between an atom and a measure supported by two points. This follows because I can decrease the number of points in the combined supports by decomposing further if I'm not in this situation yet. For example, if $$ a:=\max\textrm{supp}\,\mu_j>\max\textrm{supp}\,\nu_j=:b $$ and (let's say) $\nu_j(b)\le \mu_j(a)$, then I can represent $\nu_j(b)\delta_b$ as a convex combination of part $\mu_j$, and I have eliminated one point from the union of the supports. Similar arguments work in the other cases.

So if $\mu\sim\nu$, I can get from $\mu$ to $\nu$ by splitting and/or combining each point mass $\mu(n)\delta_n$, and for each such operation only two other points are involved. Also, since there are only finitely many other points within reach $A$, finitely many such operations suffice for a given $n$. In other words, I can partition each point mass $\mu(n)=w_1 + \ldots + w_{N_n}$ into finitely many parts, and each part $w_j$ gets a splitting or combining treatment.

Now if $\mu\sim\nu$, $\mu\sim\nu'$, then we can consider the common refinements of these partitions. We obtain corresponding finer decompositions of $\nu$, $\nu'$ (compared to the original ones), and these witness that $\nu\sim\nu'$.