In general suppose I have an object $E$ living over a field $K$ (this could be a $K$-algebra of some kind or a scheme over $K$) such that I can talk about its extension of scalars $E_L$ along a field extension $K \to L$. If $E, E'$ are two objects over $K$ that become isomorphic after an extension, they are said to be forms of each other. When $L/K$ is a Galois extension with Galois group $G$, and under some mild conditions on what is meant by "object," forms of $E$ are classified by a Galois cohomology set (not group, sadly)
$$H^1(G, \text{Aut}(E_L))$$
and in particular, objects generally have multiple forms.
Specializing now to the case of elliptic curves, the $j$-invariant is invariant under extension of scalars, in the sense that if an elliptic curve $E$ over a field $K$ has $j$-invariant $j(E) \in K$ and we then extend scalars to $E_L$ via a map $f : K \to L$, then $j(E_L) = f(j(E))$. So two elliptic curves over $K$ with the same $j$-invariant continue to have the same $j$-invariant after arbitrary extension of scalars; and general theory says this implies, hence is equivalent to, the statement that they are isomorphic over $\overline{K}$.
However, they need not be isomorphic over $K$, so could be nontrivial forms of each other. Generically (let me stick to characteristic $0$) an elliptic curve has automorphism group $\text{Aut}(E_L) \cong C_2$ given in terms of the group law by $z \mapsto -z$, and the Galois group acts trivially on this. So in this nice special case, forms are classified by
$$\text{Hom}(G, C_2)$$
and by the Galois correspondence, nontrivial homomorphisms $G \to C_2$ correspond to quadratic subextensions of $L/K$; the corresponding forms of $E$ are the quadratic twists that Noam Elkies described.
In two exceptional cases $j = 0, 1728$ there are extra automorphisms: these elliptic curves have larger automorphism groups $C_6, C_4$ corresponding to sixth resp. fourth roots of unity, and now the Galois action is nontrivial and given by the action of $G$ on roots of unity. So the Galois cohomology in this case is
$$H^1(G, \mu_6), H^1(G, \mu_4)$$
which, if we now take $L$ to be the algebraic closure, can be identified with $K^{\times}/(K^{\times})^6, K^{\times}/(K^{\times})^4$ by Kummer theory (similarly in the previous case we have $H^1(G, \mu_2) \cong K^{\times}/(K^{\times})^2$). So we get sextic and quartic twists, as Will Sawin mentioned. This can all be made more explicit in coordinates.