I was thinking of this case: two vector spaces - $\mathbb{R}^2$ over $\mathbb{R}$ and $\mathbb{C}^2$ over $\mathbb{C}$. Every basis in the first vector space is also a basis for the second vector space.
I am only starting linear algebra (also with proofs), so not sure if it makes any sense, but I will just give it a shot.
Suppose we have two vector spaces $F^n$ over $F$ and $S^n$ over $S$, then the vector space $S^n$ has the same basis as $F^n$ iff $\dim{S^n}$ = $\dim{F^n}$ and $F\subseteq S$.
This is my idea of the proof:
$V= { \{\vec{v_1}, \vec{v_2}, ..,\vec{v_n}\} }$ is some basis of $F^n$ and by the definiton $\vec{u} = (c_1\vec{v_1} + c_2\vec{v_2} + ...+c_n\vec{v_3})$, for every $\vec{u}\in F^n$ and $c_i \in F$. Because $F \subseteq S$, $\vec{u'} = (a_1\vec{v_1} + a_2\vec{v_2} + ...+a_n\vec{v_3})$ for every $\vec{u'}\in S^n$ and $a_i\in F \in S $.
I apologize if I have made any logical mistakes.