Is the determinant the "only" group homomorphism from $\mathrm{GL}_n(\mathbb R)$ to $\mathbb R^\times$?
In general, if $\phi : G \to H$ is a Lie group homomorphism then we get a Lie algebra homomorphism $\phi_* :\mathfrak g \to \mathfrak h$. If $\phi,\psi : G\to H$ are two Lie group homomorphisms then $\phi_* = \psi_*$ implies $\phi = \psi$. So we really just have to classify Lie algebra homomorphisms $\mathfrak{gl}_n(\mathbb R)\to \mathbb R$ and then see which ones lift to the groups.
Since the Lie algebra $\mathbb R$ is abelian, these Lie algebra homomorphisms are those elements of $\mathfrak{gl}_n(\mathbb R)^*$ that vanish on commutators. It turns out that this property uniquely characterizes the trace up to scale (I wasn't able to find a good reference for this but it should be easy, if a bit messy, to prove by hand).
So given $\phi : \mathrm{GL}_n(\mathbb R) \to \mathbb R^\times$, we have $\phi_* = c \operatorname{tr}$ for some $c \in \mathbb R$. Now $\phi_*$ restricts to the connected components where it integrates to $\mathrm{GL}_n(\mathbb R)_+ \ni A\mapsto(\det A)^c \in \mathbb R^+$. Of course for this homomorphism to extend to all of $\mathrm{GL}_n(\mathbb R)$ requires that the map $a \mapsto a^c$ be well-defined on all of $\mathbb R^\times$ so Lie group homomorphism is $\det$ followed by an automorphism of $\mathbb R^\times$.
Here's a more elementary proof, which has the advantage of not using the assumption that the homomorphism is continuous. (It is a very minor adaptation of the proof of Theorem 3.2 in Cullen's Matrices and Linear Transformations, an early step in Cullen's axiomatic development of the determinant.)
Let $\varphi\colon \text{GL}_n(\mathbb{R})\to\mathbb{R}_\times$ be a group homomorphism. Denote elementary matrices as follows:
- $I_{kR_i}$ is the matrix obtained by multiplying the $i$th row of the identity matrix by $k$.
- $I_{kR_i+R_j}$ is the matrix obtained by adding $k$ times the $i$ row of the identity matrix to the $j$th row.
- $I_{R_i\leftrightarrow R_j}$ is the matrix obtained by exchanging the $i$th and $j$th rows of the identity matrix.
Let $i\colon\mathbb{R}_\times\to\text{GL}_n(\mathbb{R})$ be defined by $i(x) = I_{xR_1}$. Note that $i$ is a homomorphism. We will show that, for all $A\in\text{GL}_n(\mathbb{R})$, we have $$ \varphi(A) = \varphi(i(\det(A))) \tag{$\ast$} $$ This establishes the desired result with $f=\varphi\circ i$.
First, ($\ast$) holds for $A=I_{xR_1}$, since then $A=i(\det(A))$.
Second, from the identity $$ I_{kR_j} = I_{R_1\leftrightarrow R_j} I_{kR_1} I_{R_1\leftrightarrow R_j} $$ we obtain \begin{align*} \varphi(I_{kR_j}) &= \varphi(I_{R_1\leftrightarrow R_j} I_{kR_1} I_{R_1\leftrightarrow R_j}) \\ &= \varphi(I_{R_1\leftrightarrow R_j} I_{R_1\leftrightarrow R_j} I_{kR_1}) &&\text{(since $\mathbb{R}_\times$ is abelian)} \\ &= \varphi(I_{kR_1}) \end{align*} which yields that ($\ast$) holds for elementary matrices of type 1.
Similarly, from the identity $$ I_{kR_j+R_i} = I_{k^{-1}R_j} I_{1R_j+R_i} I_{kR_j} $$ we get $\varphi(I_{kR_j+R_i}) = \varphi(I_{1R_j+R_i})$ for all $k\in\mathbb{R}$ (except $k=0$, which can be handled separately). But since $I_{1R_j+R_i}^2=I_{2R_j+R_i}$, this yields $\varphi(I_{1R_j+R_i})=1$, whence $\varphi(I_{kR_j+R_i})=1$, verifying ($\ast$) for elementary matrices of type 2.
Similarly again, from the identity $$ I_{R_i\leftrightarrow R_j} = I_{(-1)R_i} I_{1R_i+R_j} I_{(-1)R_j+R_i} I_{1R_i+R_j} $$ we get $\varphi(I_{R_i\leftrightarrow R_j}) = \varphi(i(-1))$, which establishes ($\ast$) for elementary matrices of type 3.
Finally, since elementary matrices generate $\text{GL}_n(\mathbb{R})$, this establishes ($\ast$) for all invertible matrices.