$\def\GL{\operatorname{GL}}$ Lemma Let $K$ be a field and $A\in\GL(n,K)$. The map $\gamma_A^K:C\in M_n(K)\mapsto ACA^{-1}\in M_n(K)$, which is a $K$-linear map, has determinant $\delta^K(A)=1$.
Proof. Since $\gamma_A^K\circ\gamma_B^K=\gamma_{AB}^K$, we have $\delta^K(AB)=\delta^K(A)\delta^K(B)$, and then $\delta^K(A)=\delta^K(DAD^{-1})$ for all $D$. In other words, the function $\delta^K$ is invariant under conjugation in $\GL(n,K)$.
Now, if $A$ is diagonal with diagonal elements $\lambda_1$, $\dots$, $\lambda_n$, then the map $\gamma_A$ is $(c_{i,j})\mapsto(\lambda_i\lambda_j^{-1}c_{i,j})$. We see at once at the eigenvectors are the elementary matrices and that the eigenvalues are the numbers $\lambda_i\lambda_j^{-1}$. The determinant $\delta^K(A)$, which is the product of the eigenvalues, is in this case then easily seen to be $1$. We therefore have that $\delta^K(A)=1$ for all diagonal invertible matrices, and it follows from this that $\delta^K(A)=1$ for all diagonalizable invertible matrices because of conjugation invariance.
Now, if the field is algebraically closed, then the set of diagonalizable matrices is Zariski dense in the set of all invertible matrices, and the map $\delta$ is regular —it is a rational function in the coefficients— so $\delta$ is actually identically $1$.
If $K$ is not algebraically closed, let $\bar K$ be its algebraic closure. Then of course $A\in\GL(n,\bar K)$ and by what we have proved we have $\delta^{\bar K}(A)=1$. It is easy to see that $\delta^K(A)=\delta^{\bar K}(A)$, so we reach the conclusion we want also in this case. $\Box$
Lemma. Let $L/K$ be a finite field extension and let $N:L\to K$ be the norm, so that $N(l)$ is the determinant of the $K$-linear map $x\in L\mapsto lx\in L$. If $V$ is an $L$-vector space and $f:V\to V$ is an $L$-linear map, let $\det_Lf$ and $\det_Kf$ be the determinants of $f$ viewed as an $L$-linear map and as a $K$-linear map. Then $N(\det_Lf)=\det_Kf$.
Proof. We can suppose that $f$ preserves a complete flag $0=V_0\subset V_1\subset\cdots\subset V_n=V$ of $L$-subspaces of $V$; if not, we can extend scalars to a finite extension $L'/L$ which this does happen, and check that the end result is the same.
Then $\dim_LV_i/V_{i-1}=1$ for all $1\leq i\leq n$, and $f$ induces an $L$-linear map $f_i:V_i/V_{i-1}\to V_i/V_{i-1}$ which is given by multiplication by some scalar $l_i\in L$. We have $\det_L=l_1\dots l_n$.
Now the $V_i$'s are of course also $K$-subspaces of $V$, forming now a partial flag in $V$ viewed as a $K$-vector space, and $f$ preserves it.
The determinant of $f$ viewed as a $K$-linear map is then the product of the determinants of the maps $f_i$ viewed themselves as $K$-linear maps. Since we know that $f_i$ is essentially multiplication by $l_i$ on $L$, by definition, $\det_Kf_i=N(l_i)$. The lemma follows from this. $\Box$
In this proof, we are computing the determinant of a matrix by first taking it to a block upper diagonal form, so to speak. We can now do what we really wanted:
Proposition. Let $K\subseteq L$ be a finite field extension and let $A\in\GL(n,L)$. Then the determinant of the $K$-linear map $C\in M_n(L)\mapsto ACA^{-1}\in M_n(L)$ is $1$.
Proof. By the second lemma, the determinat in question is the norm for the extension $L/K$ of the determinant of the same map viewed as an $L$-linear map. We computed this last determinant to be $1$ in the first lemma, and $N(1)=1$. $\Box$
