Assume that $K,L$ are fields such that there is an isomorphism of groups $\mathrm{GL}_n(K) \cong \mathrm{GL}_n(L)$ for all $n \in \mathbb{N}$. Does it follow that $K \cong L$?
I am also interested in related results, for example if it is enough to test small values for $n$ and what happens for $\mathrm{PGL}_n$.
Here is a proof when $\rm{char}(K)$ and $\rm{char}(L)$ are $\neq 2$.
Let $\phi : \rm{GL}_2(K) \xrightarrow{\sim} \rm{GL}_2(L)$ be an isomorphism. Then $\phi$ induces an isomorphism between the derived subgroups $\rm{SL_2}$.
Let's assume the following lemma (I may add the proof later) (we need $\rm{char}(K) \neq 2$ for this lemma).
Lemma: $\phi$ sends any transvection to a transvection.
Then $\phi(\begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix})$ and $\phi(\begin{smallmatrix} 1 & 0 \\ 1 & 1 \end{smallmatrix})$ are both transvections. Choose $e_1$ an eigen-vector of the former and $e_2$ an eigen-vector or the latter. Since the two matrices do not commute, $(e_1,e_2)$ is a basis. In this basis $\phi(\begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix})$ and $\phi(\begin{smallmatrix} 1 & 0 \\ 1 & 1 \end{smallmatrix})$ are of the form $(\begin{smallmatrix} 1 & * \\ 0 & 1 \end{smallmatrix})$ and $(\begin{smallmatrix} 1 & 0 \\ * & 1 \end{smallmatrix})$. So we can assume that $$\phi\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, \quad \phi\begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ * & 1 \end{pmatrix}.$$ The centralizer of $(\begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix})$ is $$C\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \left\{ \begin{pmatrix} a & x \\ 0 & a \end{pmatrix} ~|~ a,x \in K \text{ or } L \right\},$$ and the normalizer of $(\begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix})$ is $$N\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \left\{ \begin{pmatrix} a & x \\ 0 & b \end{pmatrix} ~|~ a,b,x \in K \text{ or } L \right\},$$ and same for $(\begin{smallmatrix} 1 & 0 \\ 1 & 1 \end{smallmatrix})$.
The subgroup $\{ (\begin{smallmatrix} 1 & x \\ 0 & 1 \end{smallmatrix}) \}$ is the set of transvections centralizing $(\begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix})$, hence there is an additive isomorphism $\sigma : K \rightarrow L$, such that $\phi(\begin{smallmatrix} 1 & x \\ 0 & 1 \end{smallmatrix}) = (\begin{smallmatrix} 1 & \sigma(x) \\ 0 & 1 \end{smallmatrix})$ and $\sigma(1)=1$. If we show that $\sigma$ is multiplicative, then we are done.
The subgroup $\{ (\begin{smallmatrix} a & 0 \\ 0 & a^{-1} \end{smallmatrix})\}$ is the set of matrices in the derived subgroup normalizing both $(\begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix})$ and $(\begin{smallmatrix} 1 & 0 \\ 1 & 1 \end{smallmatrix})$, hence there is multiplicative isomorphism $\tau : K^* \rightarrow L^*$ such that $\phi(\begin{smallmatrix} a & 0 \\ 0 & a^{-1} \end{smallmatrix}) = (\begin{smallmatrix} \tau(a) & 0 \\ 0 & \tau(a)^{-1} \end{smallmatrix})$. Note that $$\begin{pmatrix} a & 0 \\ 0 & a^{-1} \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} a & 0 \\ 0 & a^{-1} \end{pmatrix}^{-1} = \begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix}.$$ Apply $\phi$ in both sides and get $$\begin{pmatrix} \tau(a) & 0 \\ 0 & \tau(a)^{-1} \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} \tau(a) & 0 \\ 0 & \tau(a)^{-1} \end{pmatrix}^{-1} = \begin{pmatrix} 1 & \sigma(a) \\ 0 & 1 \end{pmatrix}.$$ Hence $\sigma(a) = \tau(a)$ for all $a \in K^*$, and $\sigma : K \rightarrow L$ is an isomorphism of field.
EDIT: proof of the lemma (an isomorphim sends a transvection to a transvection).
Lemma 1: Assume that $K$ (or equivalently $L$) is finite. Denote $p$ the common caracteristic of $K$ and $L$. Then $\phi$ sends a transvection to a transvection.
proof: It is because in fields of caracteristic $p$, elements of order $p$ in $GL_2(K)$ are exactly transvections. Indeed a transvection can be written $A = (\begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix})$ and $A^p = (\begin{smallmatrix} 1 & p \\ 0 & 1 \end{smallmatrix}) = I_2$, and conversely if an elment $A$ has order $p$, then the polynomial $X^p-1 = (X-1)^{p}$ cancels $A$, hence $A$ is triangular with eigenvalues equal to 1.
From now assume that $L$ and $K$ are infnite. For $G$ a group and $g \in G$, denote $C(g)$ the centralizer of $g$ and $N(C(g))$ the normalizer of $C(g)$ and $Z(G)$ the center of $G$. We say that $g$ satisfies the property (P) if :
there exists $x \in N(C(g)) \setminus C(g)$ such that $x^2 \notin Z(G)$.
Since the derived group of $GL_2$ is $SL_2$, $\phi$ induces an isomorphism $SL_2(K) \xrightarrow{\sim} SL_2(L)$.
Lemma 2: Elements of $SL_2(K)$ (or $SL_2(L)$) satisfying the property (P) are exactly the transvections and the opposite of transvections.
proof: $\bullet$ If $A \in SL_2(K)$ is a transvection, say $A = (\begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix})$, then $$ C(A) = \left\{ \begin{pmatrix} \pm 1 & * \\ 0 & \pm 1 \end{pmatrix} \right\}$$ $$ N(C(A)) = \left\{ \begin{pmatrix} a & * \\ 0 & a^{-1} \end{pmatrix} \right\}$$ Using the fact that $K$ is infinite, we can find an element $x \in N(C(A)) \setminus C(A)$ such that $x$ is not scalar (just take $x = (\begin{smallmatrix} a & 1 \\ 0 & a^{-1} \end{smallmatrix})$ with $a^2 +1 \neq 0$). Hence $A$ stisfy property (P).
$\bullet$ If $A \in SL_2(K)$ is the opposite of a transvection, say $A = (\begin{smallmatrix} -1 & 1 \\ 0 & -1 \end{smallmatrix})$, then the same calculation holds and $A$ satifies property (P).
$\bullet$ Now conversely if $A \in SL_2(K)$ satisfy the property (P). The characteristic polynomial $\chi_A$ has either $1$ root, $2$ distinct roots or is irreductible.
$-$ If $\chi_A$ has 2 distincts roots, then $A = (\begin{smallmatrix} a & 0 \\ 0 & b \end{smallmatrix})$ and $$ C(A) = \left\{ \begin{pmatrix} * & 0 \\ 0 & * \end{pmatrix} \right\}$$ $$ N(C(A)) = \left\{ \begin{pmatrix} * & 0 \\ 0 & * \end{pmatrix} \right\} \sqcup \left\{ \begin{pmatrix} 0 & * \\ * & 0 \end{pmatrix} \right\}$$ since the square of all elements in $N(C(A)) \setminus C(A)$ are scalar, $A$ does not satisfy property (P).
$-$ If $\chi_A$ is irreductible. Then $C(A) = K[A] \cap GL_2(K)$. We prove that $A$ can't satisfy (P). Let $x \in N(C(A))) \setminus C(A)$. Then $x$ induces an automorphism $f : K[A] \rightarrow K[A]$, $f(B) = x\cdot B \cdot x^{-1}$. Since $K[A]$ is a field of degree 2 over $K$, $f^2 = Id$. Hence $A$ commutes with $x^2$. Since $1$ and $x$ also commutes with $x^2$, and $x \notin C(A) = K[A] \cap GL_2(K)$, the centralizer of $x^2$ in $M_2(K)$ has at least dimension 3, which implies that $x$ is scalar.
$-$ It remains the case where $\chi_A$ has 1 root. But since $A$ has deteminant 1, this root is either 1 (and $A$ is a transvection) or -1 (and $A$ is the opposite of a transvection).
Lemma 3: If $K$ and $L$ has characteristic different from 2, then $\phi$ sends a transvection to a transvection.
proof: Let $A \in SL_2(K)$ be a transvection. Since the caracteristic of $K$ is different from 2, there exists $B \in SL_2(K)$ a transvection such that $B^2=A$. According to lemma 2, $\phi(B)$ is a transvection or the opposite of a transvection. In both cases $\phi(A) = \phi(B)^2$ is a transvection.
