Given two elliptic curves over the field $K$ by
$$Y^2=X^3+A_1x+B_1$$
and
$$Y^2=X^3+A_2x+B_2$$
Now I want to prove that two elliptic curves are isomorph over$\bar{K}$ if and only if
$A_1=u^4A_2$ and $B_1=u^6B_2$ for $u \in \bar{K^*}$, denoting the invertible algebraic closure. Because if we define $$\phi: (x,y) \rightarrow (u^2x,u^3y)$$ we get
$$Y^2=X^3+A_1x+B_1 \rightarrow u^6Y^2=u^6X^3+u^2A_1x+B_1$$ which is equivalent to $$Y^2=X^3+A_2x+B_2$$
Now in my opinion I need to show two things: First: This is indeed an isomorphism; For this reason I need to show that it preserves the group structure, so $$\phi(P_1+P_2)=\phi(P_1)+\phi(P_2)$$ for two points $P_1$ and $P_2$ on the elliptic curve; Is this true? How do I show that?
Moreover I need to show that there is no other isomorphism that means all isomorphisms are of the form $$\phi: (x,y) \rightarrow (u^2x,u^3y)$$ An idea how I can do that?
I'd be very happy about some help.
Please note: I already asked this questions 6 hours ago but didn't receive any comment or answer. To avoid a question being asked twice I deleted the old post and started this one because I wasn't really confident of getting a reply after 6 hours without any answer. I hope this is okay for everyone