Two finite-dimensional vector spaces over $F$ are linearly isomorphic if and only if they have the same dimension

If $T:V\to W$ is an isomorphism and $(b_j{)}_{j<n}$ is a basis of $V$, then $(T(b_j){)}_{j<n}$ is independent because applying $T^{-1}$ to a vanishing linear combination makes every coefficient zero, and it spans because every $w$ equals $T(v)$ and $v$ expands in the $b_j$. Thus it is a basis of $W$, so the dimensions agree.

Conversely, if the dimensions agree, choose ordered bases $(b_j{)}_{j<n}$ of $V$ and $(c_j{)}_{j<n}$ of $W$ and define $T({\sum }_j{x}_j{b}_j)={\sum }_j{x}_j{c}_j$. Unique coordinates make this a linear map with $T(b_j)=c_j$.

Defining $S({\sum }_j{y}_j{c}_j)={\sum }_j{y}_j{b}_j$ gives a linear inverse to $T$. For $n=0$, both bases are empty and both spaces are zero, so the same formulas give the unique isomorphism.