Why does the Hamiltonian define symmetry/invariance? In Sakurai's Modern Quantum Mechanics, in Chapter 4, he effectively states that the operation of rotation or translation, represented by a unitary operator $U$, is customarily called a symmetry operator regardless of whether the physical system itself possesses the symmetry corresponding to $U$. It's a symmetry or invariance of the system only when $U^\dagger H U=H$.
Why are symmetries defined with respect to invariance of the Hamiltonian?
 A: In classical mechanics, a conserved quantity has vanishing Poisson bracket with the Hamiltonian. Such quantities become "good quantum numbers" in QM: they commute with $H$, so simultaneous eigenstates from a complete basis. The evolution operator $e^{-iHt/\hbar}$ also commutes with good quantum numbers, so their probability distribution is unchanged. (Thus the connection from Noether's theorem of conservation laws to continuous symmetries survives in QM.) For unitary $U$ commuting with $H$, $U^\dagger=U^{-1}$ obtains your equation.
A: I interpret your question as "why would one want to call this the definition of symmetry?"
My imagination of symmetry: different "view" of the states such that the physics looks the same. Specifically, the states evolve in the same way. 
In quantum mechanics, the unique role played by Hamiltonian is that it's the operator of time evolution: $e^{-i H t}$. Different views correspond to the change of states, e.g., rotation every state by an angle spatially. To guarantee the states of the new view is equivalent to the original states, at least the inner products between themselves should be the same. Thus, it's realized by unitary operator $U$: $|\psi \rangle \rightarrow U |\psi \rangle$. To make sure the new states in the new "view" evolve in the same way, the evolution of a transformed state $ e^{-i H t} U |\psi \rangle$ should same as the transformed evolved state $U e^{-i H t}  |\psi \rangle$ (excluding time-reversal symmetry). 
Because the property should be true for every state, $e^{-i H t} =U^{-1} e^{-i H t} U$. Because it should also be true for any length of time evolution, $H=U^{-1} H U$.
A: In the simplest quantization, the classical Poisson bracket $\{A,B\}$ of two functions is mapped to the commutator bracket
$$
\{A,B\} \mapsto \frac{1}{i\hbar} [\hat A,\hat B] \tag{1}
$$
of two observables.  The simplest example of this is $\{x,p\}=1$ that becomes $[\hat x,\hat p]=i\hbar I$.
A symmetry is associated with a conserved quantity $Q$, which in classical mechanics implies $\{Q,H\}=0$.  For instance if the system has cylindrical symmetry, the momentum $p_\phi$ conjugate to the azimuthal angle $\phi$ is conserved in the sense above.  The connection between symmetries and conserved quantities is through Noether's theorem.   
Clearly if $\{Q,H\}=0$ is a symmetry in the classical sense, then one would expect that the quantum version would be $[\hat Q,\hat H]=0$. 
If you have $[\hat Q,\hat H]=0$, is it easy to see that, defining 
$$
U=e^{-i \alpha \hat Q}
$$
one obtains
\begin{align}
U\hat H U^{-1} &= \left(1-i\alpha \hat Q+\ldots \right)\hat H\left(1+i\alpha \hat Q+\ldots\right)\\
&= \hat H -i\alpha[\hat Q,\hat H]+\ldots  \tag{2}\\
&=H
\end{align}
where the remaining terms in (2) are all nested commutators of the type
$$
[\hat Q,[\ldots,[\hat Q,[\hat Q,\hat H]]\ldots ]
$$
and are $0$ since $[\hat Q,\hat H]=0$.
