How can I prove that the following 2 prehomogeneous vector spaces are not isomorphic?
1)$(GL\_n,\Lambda\_1\oplus \Lambda\_1,\mathbb{C}^n \oplus \mathbb{C}^n)$
2)$(GL\_n,\Lambda\_1\oplus \Lambda\_1^\*,\mathbb{C}^n \oplus \mathbb{C}^n)$
where $\Lambda\_1$ is the standard representation of $GL\_n$ on $\mathbb{C}^n$.
in the case of prehomogeneous vector spaces the notion of isomorphism is given by:


Two triplets $(G, \rho, V)$ and $(G', \rho', V')$ are isomorphic if there exist a rational isomorphism $\sigma : \rho(G) \to \rho'(G')$ and an isomorphism $\tau : V \to V'$, both defined over $\mathbb{C}$, such that $$\tau(\rho(g)x)=\sigma\rho(g)(\tau(x))$$ for all $g\in G$ and $x\in V$. That is the following diagram is commutative for all $g\in G$:


$$\require{AMScd}
\begin{CD}
V @>{\tau}>> V'\\
@V{\rho(g)}VV @VV{\sigma\rho(g)}V \\
V @>{\tau}>> V'
\end{CD}
$$



Prehomogeneous vector spaces
Your two representations of $GL\_n$ are not isomorphic, because one of them contains $\Lambda\_1$ with multiplicity 2 and the other with multiplicity 1, $\Lambda\_1^\*$ and $\Lambda\_1$ being non-isomorphic. More generally, if $V$ and $W$ are finite-dimensional rational representations of $G=GL\_n$ with $V$ irreducible then the multiplicity of $V$ in $W,$ $$\dim\operatorname{Hom}\_{G}(V,W),$$ is an invariant of $W$ modulo isomorphism. 


By the way, the second space is not prehomogeneous, because the bilinear pairing between $\Lambda\_1$ and $\Lambda\_1^\*$ is a non-constant polynomial $GL\_n$-invariant. 


