Is entropy change of the system for reversible process same as entropy change of the system for irreversible change?
Being a state function, entropy is independent of path, so the entropy change of the system is indeed the same for both a reversible and an irreversible process. The transferred heat, however, is not, and does depend on process. 

You've probably heard mentioned that the entropy change through a reversible process is less than that through an irreversible process. At first glance this proposition seems suspicious, because we've just said that as a state function entropy is independent of path$-$the resolution is that the entropy change of the system remains the same regardless of process, but the entropy change of the environment is indeed different, such that the total entropy change is greater for an irreversible process than for a reversible process. [Interpreted differently: the environment is in a different state after the system undergoes a reversible process than after it undergoes an irreversible process.]

The internal energy does depend on the system's entropy$-$by the first law we may write $$\mathrm{d}U = T\,\mathrm{d}S + \mathbf{f\cdot\mathrm{d}X}$$ (where $\mathbf{f\cdot\mathrm{d}X}$ represents a generalized work term; commonly this is $-p\,\mathrm{d}V$), so we see that $U$ is a natural function of $S$ and $\mathbf{X}$; $U = U(S,\mathbf{X})$. To give a functional form for $U$ as a function of $S$ and $\mathbf{X}$, we'll need in general an equation of state, an equation for energy, and a type of process$-$let's restrict ourselves to the case of an ideal gas, for which the two equations are $$pV = nRT \quad \text{and} \quad U = \frac{3}{2}nRT$$ (assuming also a monatomic gas), and work out how $U$ varies over an isobaric process. We want to evaluate $\mathrm{d}U = T\,\mathrm{d}S - p\,\mathrm{d}V$; let's start by working out $-p\,\mathrm{d}V$: $$-p\,\mathrm{d}V = -p\,\mathrm{d}\left(\frac{nRT}{p}\right) = -nR\,\mathrm{d}T = -\frac{2}{3}\,\mathrm{d}U.$$ Now plug that into our starting equation, substitute, and rearrange... $$\frac{5}{3}\,\mathrm{d}U = T\,\mathrm{d}S = \frac{2U}{3nR}\,\mathrm{d}S \quad \Longleftrightarrow \quad \frac{5}{2}nR\,\frac{\mathrm{d}U}{U} = \mathrm{d}S,$$ which integrates to $$\Delta S = \frac{5}{2}nR\ln\left(\frac{U_2}{U_1}\right) = \frac{5}{2}nR\ln\left(\frac{T_2}{T_1}\right).$$ This is no different from working out $q_\text{rev}$ for such a process and then using $\Delta S = q_\text{rev}/T$; the role of energy is simply more explicit in this approach.