= Solution
Use the convention that a <sesquilinear form> is linear in its first argument and conjugate-linear in its second: for complex scalars $a,b$,
$$
\phi(av_1+bv_2,w)=a\phi(v_1,w)+b\phi(v_2,w),\qquad \phi(v,aw_1+bw_2)=\bar a\phi(v,w_1)+\bar b\phi(v,w_2).
$$
No symmetry or positivity is assumed. Put $\chi=\phi-\psi$. Its diagonal vanishes. Expansion at $v+w$ gives
$$
\chi(v,w)+\chi(w,v)=0,
$$
while expansion at $v+iw$ gives
$$
-i\chi(v,w)+i\chi(w,v)=0.
$$
These two equations force $\chi(v,w)=\chi(w,v)=0$. This is the <polarization argument for a vanishing quadratic form>, and proves \b[$\phi=\psi$ on every pair of vectors].
For the final deduction, the <linear map> $\alpha$ makes $\psi(v,w)=\phi(\alpha v,\alpha w)$ another <sesquilinear form>. Its diagonal agrees with that of $\phi$, so the result just proved gives
$$
\boxed{\phi(\alpha v,\alpha w)=\phi(v,w)\quad\text{for all }v,w\in V}.
$$
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