A proof by induction (on the dimension $n$ of the vector-space $X\simeq K^n$ or $\mathbb{C}^n$).

1) The basis step, $n=1$, is trivial.

2) 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. Let $\{x_j\}_{0\leq j \leq n}$ be a basis of $X$ and $\{x_j^*\}_{0\leq j \leq n}$ a dual basis.

3) $T$ is injective and preserves rank. For rank 1 matrices, one of the implications is that
$$\forall j \in \{0,\ldots,n\}:T(x_0x_j^*) = y_0 \varphi_j^* \text{ and }T(x_jx_0^*)=y_j\varphi_0^*,\\
\text{or},
\forall j \in \{0,\ldots,n\}:T(x_0x_j^*) = y_j \varphi_0^*\text{ and }T(x_jx_0^*)=y_0\varphi_j^* \qquad(1)$$
where $\{y_j\}_{0\leq j \leq n}$ is a basis of $X$ ($\{y_j^*\}_j$ the corresponding dual basis) and $\{\varphi_j^*\}_{0\leq j \leq n}$ is a basis of $X^*$ ($\{\varphi_j\}_j$ the corresponding dual basis, keeping $X^{**}\simeq X$ in mind). We can restrict ourselves to the former scenario outlined in (1) by replacing $T$ in the latter scenario with $T\circ t$, where $t$ is a "coordinate-transpose", i.e. $t(x_jx_k^*):=x_kx_j^*$ and $t$ linear. If we then define the invertible matrices $U_1=\sum_{j=0}^n x_j y_j^*,\,V_1=\sum_{j=0}^n \varphi_jx_j^*$, then $T_1(.):=U_1T(.)V_1$ fixes both $\{x_0x_j^*\}_j$ and $\{x_jx_0^*\}_j$, i.e. $T_1$ fixes matrices whose non-zero entries occur only in the zeroth row and/or zeroth column. (Note that $T_1$ multiplies the determinant of its input by an uncertain factor $\lambda:=\det(U_1 V_1)\neq 0$)

4) 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 $\forall A \in \End(W)$
$$\lambda\det(A) = \lambda\det(x_0x_0^*+i\circ A \circ \pi)=\det(T_1(x_0x_0^*+i\circ A \circ \pi))=\det(x_0x_0^*+T_1(i\circ A \circ \pi))=\det(\tilde{T}(A)).$$
So according to the induction hypothesis there exist $U_2$, $V_2$ with $\det(U_2V_2)=\lambda$ s.t. $\tilde{T}(.) \equiv U_2(.)V_2$ or $\tilde{T}(.) \equiv U_2(t(.))V_2$.

5) It is now easy to check (using only the property that the below-defined $T_2$ preserves rank 1 matrices) that
$$T_2(.):=(x_0x_0^* + i\circ U_2^{-1}\circ \pi) *T_1(.)*(x_0x_0^* +i\circ V_2^{-1}\circ \pi)$$
is the identity map on $\End(X)$. (The latter scenario that was hypothesized at the end of step 4) is found to be impossible)