Why does a catalyst increase the forward and reverse reactions equally
The equilibrium is unaffected because the barrier to both forwards and back reaction is reduced, and even though the amount is constant it works out that this effect cancels out. This is a consequence of the fact that the population of energy levels is exponentially distributed from the Boltzmann distribution. This also leads to Arrhenius type rate expressions. 

If the energy in your figure from reactants to transition state is $E_a$ and the difference in products to reactants $G$ and we lower the energy $E_a$ by $E_0$ then using Arrhenius type rate expressions the equilibrium constant is $K_e=k_{forward}/k_{back}$

$\displaystyle K_e=\frac{\exp(-E_a/RT)}{\exp(-(E_a+G)/RT)}$,  and for the catalyst  $\displaystyle K_e^C= \frac{\exp(-(E_a-E_0)/RT)}{\exp(-(E_a-E_0+G)/RT)}$  which is the same as $K_e$if you expand the terms in the exponentials.