Setup: (see commutative diagram below) Suppose $E = \{\sum_j^n m_{ji}x_j\}_i^n$ is a basis for the free $R$-module $F$. Let $I = \{1, \dots, n\}$ be an indexing set and let $V = \{v_1, \dots, v_n\}$ be a basis for the free $R$-module $R^{\oplus n}$. Let $\psi:I \to R^{\oplus n}$ be the canonical injection $i \mapsto v_i$. By definition of a free module basis $E$, $g$ is an injection into $F$ to the ith element of $E$, and $\varphi_E: R^{\oplus n} \to F$ is an isomorphism $\sum_i^n k_iv_i \mapsto \sum_i^n k_ig(i)$, so that the left hand side of the diagram below commutes.
I want to prove: if we define $h: I \to R^{\oplus n}: j \mapsto \sum_i^n m_{ij}v_i$, the resulting homomorphism $\phi: R^{\oplus n} \to R^{\oplus n}: \sum k_iv_i \mapsto \sum k_i h(i)$ that makes the diagram below commute is an isomorphism (i.e., $S = \{\sum^n_{i}m_{ij}v_i\}_i^n$ is a basis for $ R^{\oplus n}$)
Argument: since $\varphi_E$ is an isomorphism, the restriction of $\phi_E$ to the set $S = \{\sum^n_{i}m_{ij}v_i\}_i^n$ produces a bijection of sets with the basis set $E$, and therefore $S$ has the same cardinality as $E$, from which we obtain that the restriction of $\varphi$ to the set $V$ produces a bijection $v_i \mapsto \sum m_{ij}v_i$ with $S$. Since $\varphi$ is an extension of this bijection with a basis, it follows it is a free module isomorphism.
Question: How exactly do I justify the very last step, "since $\varphi$ is an extension of a bijection with a basis, it follows it is a free module isomorphism"? Would there be some sort of argument from category theory I could use here? This seems to me to be related to the idea that a linear map is fully determined by its action on a basis.
Define the homomorphism $f:F \to R^n:x_i \mapsto v_i$. Clearly it is an isomorphism. Now, notice:
$$\begin{align} \phi_E(v_j) &= \sum_i^n m_{ij}x_i \\ f \circ \phi_E(v_j) &= \sum_i^n m_{ij}v_i \\ f \circ \phi_E(v_j) &= \phi(v_j) \end{align}$$
Since $f, \phi_E$ are isomorphisms, so will $\phi$ be.