Here $X$ is some measurable space and $U(L^2(X,\mathbb{C}^n))$ denotes the Banach space of unitary automorphisms of the Hilbert space $L^2(X,\mathbb{C}^n)$.
Let $U(\mathbb{C},n)$ denote the unitary matrices $n\times n$, we can consider the space of $L^2$ functions $\phi: X\to U(\mathbb{C},n)$, denoted as $L^2(X, U(\mathbb{C},n))$. An element $\phi$ in the latter space defines an element in $U(L^2(X,\mathbb{C}^n))$ that sends $f\in L^2(X, \mathbb{C}^n)$ to $\phi\cdot f $ such that
$\phi \cdot f(x) = \phi(x) (f(x))$.
Since $\phi(x)$ is unitary, it preserves the norms hence $\phi\cdot f\in L^2(X,\mathbb{C}^n)$ (this is why I am taking the unitary group and not $GL$) and the operator norm is $1$. Therefore we have an inclusion
$L^2(X, U(\mathbb{C},n))\to U(L^2(X,\mathbb{C}^n)) $
Is this map an isomorphism? Can you provide an example of an element of $U(L^2(X,\mathbb{C}^n))$ which is not in the image of the above map?
I believe the answer to the first question is negative. The LHS is an Hilbert space while the RHS is a Banach (but I cannot disprove that it can't be given an Hilbert structure, e.g. is it reflexive? I don't know).
Also, I thought that $U(L^2(X,\mathbb{C}^n))$ is contractible (at least if $\dim L^2(X,\mathbb{C}^n) = \infty$) by Kuipers' theorem, therefore another possible way of disproving it would be show that $L^2(X, U(\mathbb{C},n))$ has more connected components.
There are several problems here. First, you need a measure to define $L^2(X,U(n))$, not just a measurable space. Moreover, $L^2(X,U(n))$ is not a Hilbert space or even a vector space - neither sums nor scalar multiples of unitaries are unitary in general. In fact, if the underlying measure is infinite, then $L^2(X,U(n))=\emptyset$. For similar reasons, $U(L^2(X,\mathbb{C}^n))$ is not a vector space.
Even if the underlying measure is finie, the map from the question is almost never surjective. To see this, note that a bounded linear operator of the form $$ Tf(x)=\psi(x)f(x) $$ with $\psi(x)\in\mathbb{C}$ for a.e. $x$ commutes with all elements of the image of $L^2(X,U(n))$, but the only operators commuting with all elements of $U(L^2(X,\mathbb{C}^n))$ are scalar multiples of the identity. So the map cannot be surjective if $\dim L^2(X)\geq 2$.