Let $K$ be a finite field, and $\overline{K}$ its algebraic closure. It is well known that two curves are isomorphic over $\overline{K}$ if and only if they have the same $j$-invariant. If two such curves are also $K$-isogenous, I believe we can conclude that they are $K$-isomorphic, but I cannot find any reference or elementary proof of this fact; is it easy to see, or does anyone have a reference? (it seems that this result is implicitly used in the algorithms for isogeny graphs of elliptic curves, like Kohel's, where the isogenous curves are encoded by there $j$-invariant).


As a side question, given a $j \in K$, how many curves over $K$ have this $j$-invariant, up to $K$-isomorphisms? For example, a (non-supersingular) curve and its quadratic twist both have the same $j$-invariant, but are not $K$-isogenous, so not $K$-isomorphic, so we have at least two $K$-isomorphism classes; is this it?



$j$-invariants of elliptic curves over finite fields
As Chris and Rene said, your second question is all about twists. See specifically *The Arithmetic of Elliptic Curves* (Springer), Proposition X.5.4, which says that if the characteristic of $K$ is not 2 or 3, then the twists of $E/K$ are in one-to-one correspondence with $K^\*/(K^\*)^n$, where
$$ n=\begin{cases} 2&\text{if $j(E)\ne0,1728$},\\4&\text{if $j(E)=1728$},\\6&\text{if $j(E)=0$}.\\\end{cases}$$


So for a finite field $K=\mathbb{F}\_q$ of characteristic $p\ge5$, in the first case there are always exactly two distinct twists, but for the latter two cases, the number of twists depends on whether $K$ has a fourth root, respectively cube root, of one, or equivalently, on the value of $q\bmod4$, respectively $q\bmod3$.


