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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 $\lambda^2:=\det(U_1 V_1)\neq 0$)

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 $\det(U_2)=\det(V_2)=\lambda$ ($\lambda$ is the square root of $\lambda^2$ closest to the $\mathbb{R}^+$ axis) 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 and, if I'm not mistaken, you should find that our adjustments in step 3) have already forced $U_2=I=V_2$ a posteriori)

Another proof

1. Suppose $X$ is an $n$-dimensional $K$-vectorspace and $T:\DeclareMathOperator\End{End}\End(X) \to \End(X)$ preserves the determinant. Let $\{x_j\}_{1\leq j \leq n}$ be a basis of $X$ and $\{x_j^*\}_{1\leq j \leq n}$ its dual basis.

2. $T$ is injective and preserves rank. For rank 1 matrices, one of the implications is that $$\forall j \in \{1,\ldots,n\}:T(x_1x_j^*) = y_1 \varphi_j^* \text{ and }T(x_jx_1^*)=y_j\varphi_1^*,\\ \text{or}, \forall j \in \{1,\ldots,n\}:T(x_1x_j^*) = y_j \varphi_1^*\text{ and }T(x_jx_1^*)=y_1\varphi_j^* \qquad(1)$$ where $\{y_j\}_{1\leq j \leq n}$ is a basis of $X$ ($\{y_j^*\}_j$ the corresponding dual basis) and $\{\varphi_j^*\}_{1\leq j \leq n}$ 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 $U_1=\sum_{j=1}^n x_j y_j^*,\,V_1=\sum_{j=1}^n \varphi_jx_j^*$, then $T_1(.):=U_1T(.)V_1$ fixes both $\{x_1x_j^*\}_j$ and $\{x_jx_1^*\}_j$, i.e. $T_1$ fixes matrices whose non-zero entries occur only in the 1st row and/or first column. (Note that $T_1$ multiplies the determinant of its input by an uncertain factor $\det(U_1 V_1)\neq 0$, but later on the proof will indirectly show that this factor must be 1 anyway)

3. $T_1$ maps rank 1 matrices to rank 1 matrices. At this stage, this implies that $\forall j,k \in \{2,\ldots,n\}:\,\exists a_{jk}\in \mathbb{C}:\,T_1(x_jx_k^*)=a_{jk}x_jx_k^*$ (test $T_1$ on $(x_1+x_j)x_k^*$ and on $x_j(x_1^*+x_k^*)$ to arrive at this conclusion), i.e. $T_1$ performs a 'pointwise' multiplication on the matrix entries of its input (viewed in the $\{x_j\}_j$-basis). To see that the $a$-coefficients all have to be equal to 1, test $T_1$ once more on the rank 1 matrix $(x_1+x_j)(x_1^*+x_k^*)$. So we conclude that $T_1$ is the identity map on $\End{X}$ and retracing our steps, we see that $T(.)\equiv U_1^{-1}(.)V_1^{-1}$ or $T(.)\equiv U_1^{-1}(t(.))V_1^{-1}$.

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)

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 $\lambda^2:=\det(U_1 V_1)\neq 0$)

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 $\det(U_2)=\det(V_2)=\lambda$ ($\lambda$ is the square root of $\lambda^2$ closest to the $\mathbb{R}^+$ axis) 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 and, if I'm not mistaken, you should find that our adjustments in step 3) have already forced $U_2=I=V_2$ a posteriori)

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.

3. $T$ 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)

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 $\forall A \in \End(W)$ $$\det(A) = \det(i\circ A \circ \pi + p)=\det(T_1(i\circ A \circ \pi + p))=\det(\tilde{T}(A)).$$ 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 $\tilde{T}(.) \equiv U_2 *(.)^t*V_2$. If necessary 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)

5. 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)

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)