The two sets $\{1,2,3\}$ and $\{1,1,2,3\}$ are equal, but the two multisets $<1,2,3>$ and $<1,1,2,3>$ are not equal.
Multisets are generalizations of sets that allow repeated elements. A natural mapping exists from sets to multisets: we can write $\{1,2,3\} = <1,2,3>$ since these two objects have identical properties in all respects. Likewise, $\{1,1,2,3\} = <1,1,2,3>$
So we have $$<1,1,2,3> =\{1,1,2,3\} = \{1,2,3\} = <1,2,3>$$ But we noted that $<1,1,2,3> \neq <1,2,3>$, so we have a contradiction
What am I misunderstanding?
There is a natural mapping from sets to multisets, but not all multisets are images of sets under this mapping. The mapping takes $\{1,2,3\}$ to the multiset $\{\!\{1,2,3\}\!\}$ in which each of the three elements occurs once, but it also takes the set $\{1,1,2,3\}$ to $\{\!\{1,2,3\}\!\}$, because the set $\{1,1,2,3\}$ also contains each of $1$, $2$, and $3$ exactly once: it is equal to the set $\{1,2,3\}$. The multiset $\{\!\{1,1,2,3\}\!\}$ is not the image of any ordinary set, because no ordinary set has two distinguishable copies of a single element.
A set is completely determined by its members, so if $A$ is a set, then (for instance) the number $1$ either is or is not a member of $A$: it can’t be a member twice or three times. Multisets are specifically designed to overcome this limitation: in a multiset we can have multiple instances of a single element, but when we do, we have a multiset that is not the natural counterpart of an ordinary set.
Thus, your example should read that the set $\{1,1,2,3\}=\{1,2,3\}$ corresponds to (not is equal to) the multiset $\{\!\{1,2,3\}\!\}$.