Is the Hamiltonian some sort of connection/gauge field? I'm not sure if this is a well-defined question, but I was just looking through some old notes and noticed that the Hamiltonian in usual QM has a similar transformation as gauge fields in QFT: under time-dependent unitary transformations $|\psi'\rangle = U(t) |\psi\rangle$ the Hamiltonian transforms as
\begin{align*}
H' = U H U^\dagger + i (\partial_tU) U^\dagger
\end{align*}
Similarly under the gauge transformation $\phi'(x) = U(x) \phi(x)$ a non-abelian gauge field transforms as
\begin{align*}
A_\mu' = U A_\mu U^\dagger + \frac{i}{g}(\partial_\mu U) U^\dagger
\end{align*}
Is this just a coincidence or is there some deeper underlying reasons why these two objects should transform in the same way?
 A: 
the Hamiltonian transforms as
\begin{align*}
H' = U H U^\dagger + i (\partial_tU) U^\dagger
\end{align*}
Similarly under the gauge transformation $\phi'(x) = U(x) \phi(x)$ a non-abelian gauge field transforms as
\begin{align*}
A_\mu' = U A_\mu U^\dagger + \frac{i}{g}(\partial_\mu U) U^\dagger
\end{align*}


Is this just a coincidence or is there some deeper underlying reasons why these two objects should transform in the same way?

As an initial (somewhat pedantic) matter, I note that these two object do not transform in the same way. They transform in a similar way. That is, the symbols involved in the transformation look similar if you ignore the $g$ and the $\mu$s.
The vector potential is a spacetime four-vector:
$$
A^\mu = (A^0, \vec A)\;,
$$
where $A^0\equiv \Phi$ is the electrostatic potential (or "colorstatic" potential--or whatever--for your non-abelian field). This electrostatic potential transforms, per your equations, as:
$$
\Phi \to U\Phi U^\dagger + \frac{i}{g}\frac{\partial U}{\partial t}U^\dagger\;.
$$
Therefore, assuming your $g$ is the charge, the potential energy part transforms as:
$$
g\Phi \to Ug\Phi U^\dagger + i\frac{\partial U}{\partial t}U^\dagger\;,
$$
which is the same transformation as the Hamiltonian.
This is not unexpected since both are the zeroth component of a spacetime vector:
$$
P^\mu = (H, \vec P)
$$
and:
$$
A^\mu = (A^0, \vec A)\;.
$$
Further, in the Coulomb gauge, we know that the Hamiltonian looks like:
$$
H = \ldots + g\Phi\;,
$$
so we might expect $H$ and $\Phi$ to transform similarly.
