Two (undirected) multiplication graphs $n$, $m$ for $\mathbb{Z}/p\mathbb{Z}$ look the same ($n \sim_p m$) when
$$a n= b\operatorname{mod} p \ \ \text{ iff }\ \ b m = a \operatorname{mod} p$$
for all $a,b < p$. This is the case for $\mathbb{Z}/11\mathbb{Z}$ and
$n =2, m = 6$
$n =3, m = 4$
$n =5, m = 9$
$n =7, m = 8$
For $\mathbb{Z}/29\mathbb{Z}$ we find
$2 \sim_{29} 15$
$3 \sim_{29} 10$
$4 \sim_{29} 22$
$5 \sim_{29} 6$
$7 \sim_{29} 25$
$8 \sim_{29} 11$
$9 \sim_{29} 13$
$12 \sim_{29} 17$
$14 \sim_{29} 27$
$16 \sim_{29} 20$
$18 \sim_{29} 21$
$19 \sim_{29} 26$
$23 \sim_{29} 24$
Some observations:
$p-1 \not\sim_p n$ for all $n < p-1$
For both $p=11$ and $p=29$ all $n,m < p-1$ come in pairs. Maybe for all $p$? Or possibly larger equivalence classes for ever larger $p$? (Only even tupels, only $2^k$-tupels?)
For both $p=11$ and $p=29$ there are two consecutive pairs: $3,4$ and $7,8$ resp. $5,6$ and $23,24$. The first located more close to $1$, the second somehow symmetric and more close to $p-1$. Maybe for all $p$? Or possibly more than two for ever larger $p$? Or longer consecutive sequences?
My question is:
What's the general pattern? Which $n,m < p$ are equivalent in the sense above, i.e. $n \sim_p m$?
Consider your definition of $n \sim_p m$: $$a\cdot n \bmod p = b\ \ \text{ iff }\ \ b\cdot m \bmod p = a$$
Take $a=m$. Then $$m\cdot n \bmod p = b\ \ \text{ iff }\ \ b\cdot m \bmod p = m$$ The second equation holds iff $b \equiv 1 \bmod p$ and then the first equation gives $mn \equiv 1 \bmod p$.
Bottom line: $$ n \sim_p m \ \text{ iff }\ mn \equiv 1 \bmod p $$