Let $V$ be a finite dimensional $K$-vector space. Let $S_1,S_2\subset V$ be subspaces such that $\dim S_1=\dim S_2=n$. Show there is an automorphism $\varphi:V\to V$ such that $\varphi(S_1)=S_2$.
Let $\{a_1,\ldots,a_n\}$ be a basis for $S_1$, and let $\{b_1,\ldots,b_n\}$ be a basis for $S_2$. Extend these to bases for $V$, namely $\{a_1,\ldots,a_k\}$ and $\{b_1,\ldots,b_k\}$ where $k=\dim V$. Now define $\varphi:V\to V$ by mapping $a_i\mapsto b_i$ for each $1\leq i\leq k$ (making sure that the basis elements of $S_1$ map to the basis elements of $S_2$). This defines a linear map, and since the $b_j$ span all of $V$, $\dim\text{im }\varphi=\dim V$. Therefore $\varphi$ is surjective, hence $\varphi$ is an automorphism on $V$. Finally, it is clear that $\varphi(S_1)=S_2$ by construction. Is this correct?
That looks good to me.
Just a small comment. You say Therefore $\varphi$ is surjective, hence $\varphi$ is an automorphism on $V$. This is true.
However, you're referring to an argument which is usually only valid for finite dimensional vector spaces. This is fine as it is the case in your example. But it's not necessary. A linear map that maps a basis onto a basis is an automorphism whatever the dimension of the vector space considered is.