a basic univalence puzzle concerning the sup function on sets if sup({1,2,3})=3, why can't I use univalence and the fact that {1,2,3}≅{4,5,6}  to establish that sup({4,5,6})=3?
 A: You can use univalence like that, but then you have to properly apply it everywhere.
Suppose $e : \{1, 2,3\} \to \{4, 5, 6\}$ is an equivalence. By Univalence axiom there is a path $\mathsf{ua}(e) : \{1, 2,3\} = \{4, 5, 6\}$. There is a path
$$p : \sup \{1, 2, 3\} = 3.$$
In order to apply univalence to $\{1, 2, 3\}$ in this sitution, we have to transport $\{1, 2, 3\}$ as well as $\sup$, $p$ and the element $3$ along the path $\mathsf{ua}(e)$. What we get is something like
$$\mathsf{ua}(e) \cdot p : (e \circ \mathsf{sup} \circ e^{-1}) \{4, 5, 6\} = e(3).$$
I am not trying to be precise here, because we are already quite sloppy by writing set-theoretic stuff like $\{1, 2, 3\}$ and $\{4, 5, 6\}$, but I hope you get the idea: You have to explicitly transport the entire statement along the equivalence.
Similarly, if your wife sends you to a store to buy $\{\mathrm{bread}, \mathrm{milk}, \mathrm{eggs}\}$ and you brougth back $\{\mathrm{chococalate}, \mathrm{whiskey}, \mathrm{beer}\}$ then in order to make your wife happy, you would have to select an equivalence $e$ between these two sets, and transport the entire statement

Wife ordered $\{\mathrm{bread}, \mathrm{milk}, \mathrm{eggs}\}$.

along $e$. You would end up with something like

$e(\text{Wife})$ ordered $\{\mathrm{chococalate}, \mathrm{whiskey}, \mathrm{beer}\}$.

Are you quite sure you want to transport your wife like that?
A: Univalence is about equivalence of types inhabiting a Universe, but {1,2,3} is not such a type, it is a term of a set inhabiting U.
A: The question may reveal a confusion between univalence on one hand and transporting along an identity on the other hand. I will try to disentangle them.
Start with two sets, $\textit{i.e}$ two elements of the type hSet, let say $\lbrace 1,2,3\rbrace$, $\lbrace 4, 5, 6\rbrace$ : hSet, then recall that the type hSet is the dependent sum type $\sum X: U, \text{isaset}\,X$ (where $U$ is a type-theoretic universe and $\text{isaset}\,X$ is the proposition that the type $X$ is a set), one can prove that the type $\lbrace 1, 2, 3\rbrace =_{hSet} \lbrace 4,5,6\rbrace$ is equivalent to the type 
$\sum p : \text{pr1} \lbrace 1,2,3\rbrace =_U \text{pr1} \lbrace 4,5,6\rbrace, \text{transport}\,p\,(\text{pr2} \lbrace 1,2,3\rbrace) = \text{pr2}\lbrace 4,5,6\rbrace$. 
But $\text{isaset}(\text{pr1}\lbrace 4,5,6\rbrace)$ being a proposition, an element of the sum type above amounts to just an element of the type $\text{pr1} \lbrace 1,2,3\rbrace =_U \text{pr1} \lbrace 4,5,6\rbrace$. It is where the univalence comes into play, telling you that it's enough (in order to get an identity between our given sets) to have an equivalence between the first projections, namely an element of the type $\text{pr1}\lbrace 1,2,3\rbrace \cong \text{pr1} \lbrace 4,5,6\rbrace$.
Once you have an equality between $\lbrace 1,2,3\rbrace$ and $\lbrace 4,5,6\rbrace$ (and we have seen above what such an equality means), you can use this equality to transport any property from $\lbrace 1,2,3\rbrace$ to $\lbrace 4,5,6\rbrace$. Indeed, given a family of small types $B(X)$, with $X:hSet$, one has an element $\text{transport}_B$ of type 
$B(\lbrace 1,2,3\rbrace)\times (\lbrace 1,2,3\rbrace =_{hSet} \lbrace 4,5,6\rbrace) \rightarrow B(\lbrace 4,5,6\rbrace)$. 
Take for instance "$X$ has a sup" for $B(X)$, then from your equality (possibly resulting from the use of the univalence axiom as sketched above) and the fact that $\lbrace 1,2,3\rbrace$ has a sup one concludes, using $\text{transport}_B$, that $\lbrace 4,5,6\rbrace$ has a sup. You could take $B(X,x) := (\text{sup} X =_{\text{pr1}X} x)$ as a family of types indexed by the type ($\sum X: hSet, \text{pr1}\,X$) and try to transport along an adequate equality, but this family of types is not even well defined since the element $\text{sup}\,X$ does not always exist. I hope it illustrates how one can use, or misuse for that matter, transport along an identity. So you really need to be careful and work out the details .
A: Since this is tagged as category-theory, a resolution in terms of category theory may be informative.
If we use $\{ 1, 2, 3 \}$ and $\{ 4, 5, 6\}$ to denote objects of the category Set, then they are indeed isomorphic.
If we use $\{ 1, 2, 3 \}$ and $\{ 4, 5, 6\}$ to denote subsets of $\mathbb{N}$ (i.e. objects of the category $\mathrm{Sub}(\mathbb{N})$), then they are not isomorphic.
$\sup$ is a (partial) operation on $\mathrm{Sub}(\mathbb{N})$, so it is the latter notion of isomorphism that is relevant, not the former.
