A proof by induction (on the dimension of the vector-space $n$).
The basis step, $n=1$, is trivial.
Suppose we have proceeded up to and including dimension $n$ and suppose $V$ is $(n+1)$-dimensional and $T:\text{End}(V) \to \text{End}(V)$ preserves the determinant.
$T$ preserves rank. We are 'allowed' to suitably redefine $T_1(.) := U_1 *T(.) *V_1$, $\det(U_1 *V_1)=1$, after which we may assume that $T_1$ fixes $\text{diag}(0,...,0,1)$, i.e. the projector matrix with a single 1-entry in the bottom right corner.
Let $W\subseteq V$ be the $n$-dimensional subspace associated to the first $n$ rows or columns in the earlier-used matrix representation. Let $\tilde{T}:\text{End}(W) \to \text{End}(W): A \mapsto [T_1(A\oplus[0])]_W$ where $[...]_W$ is the restriction from $\text{End}(V)$ to $\text{End}(W)$ obtained by removing the final row and column in the matrix-representation. Using the Laplace expansion for the determinant on the $n+1$'th row or column, and using the fact that $T_1$ sends singular matrices to singular matrices, we see that $\forall A \in \text{End}(W)$ $$\det(A) = \det(A\oplus[1])=\det(T_1(A\oplus[1]))=\det(\tilde{T}(A)).$$ So according to the induction hypothesis there exists $U_2, V_2$ with $\det(U_2*V_2)=1$ s.t. $\tilde{T}(.) \equiv U_2 *(.)*V_2$ or $\tilde{T}(.) \equiv U_2 *(.)^T*V_2$. By composing $T$ with the transpose, we may assume from now on that the former scenario is the case.
It is now easy to check (using only the property that $T_1$ preserves rank and by 'testing' $T_2$ on matrices with a single non-zero entry in the final row and a single non-zero entry in the final column) that for a suitable $z\in \mathbb{C}$, $$T_2(.):=\text{diag}(1,...,1,z) *(U_2^{-1}\oplus [1]) *T_1(.)*(V_2^{-1}\oplus [1])*\text{diag}(1,...,1,z^{-1})$$ is the identity map on $\text{End}[V]$.