A proof by induction (on the dimension $n$ of the vector-space $X\simeq K^n$ or $\mathbb{C}^n$).
The basis step, $n=1$, is trivial.
Suppose we have proceeded up to and including dimension $n$ and suppose $X$ is $(n+1)$-dimensional and $T:\DeclareMathOperator\End{End}\End(X) \to \End(X)$ 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 $\DeclareMathOperator\diag{diag}\diag(1,0,\dotsc,0)=:p$, i.e. the projector matrix with a single 1-entry in the top left corner. (We will from now on maintain the basis $\{x_0,x_1,\ldots,x_{n}\}$ which is implicitly used in this matrix representation)
Let $W=\text{span}(x_1,\ldots,x_n)\subseteq X$, $\pi:X\to W: c_0x_0+c_1x_1+\ldots +c_nx_n\mapsto c_1x_1+\ldots +c_nx_n$ and $i: W \to X: w \mapsto w$. Let $\tilde{T}:\End(W) \to \End(W): A \mapsto \pi \circ T_1(i\circ A\circ \pi) \circ i$. Using the Laplace expansion for the determinant on the zeroth row or column, and using the fact that $T_1$ sends singular matrices to singular matrices, we see that $\forall A \in \End(W)$ $$\det(A) = \det(i\circ A \circ \pi + p)=\det(T_1(i\circ A \circ \pi + p))=\det(\tilde{T}(A)).$$ So according to the induction hypothesis there exist $U_2$, $V_2$ 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. (In case $n+1=2$ the question of whether or not to take a transpose is trivial at this point)
It is now easy to check (using only the property that $T_2$ preserves rank and by 'testing' $\det\circ T_2$ on matrices with a single non-zero entry in the zeroth row and a single non-zero entry in the zeroth column) that for a suitable $z\in \mathbb{C}$, $$T_2(.):=(zp + i\circ U_2^{-1}\circ \pi) *T_1(.)*(z^{-1}p +i\circ V_2^{-1}\circ \pi)$$ is the identity map on $\End(X)$. (In case $n+1=2$ we may need to take a transpose at THIS stage to correctly finish the proof)