Speed of reaction vs equilibrium
The equilibrium constant $K$ is the quotient of the rate constants of forward ($k_1$) and backward reactions ($k_{-1}$) (for elementary reactions).

E.g., for an  equilibrium
$$\ce{A + B &lt;=&gt; C + D}$$
we get
$$\frac{[C][D]}{[A][B]} = K = \frac{k_1}{k_{-1}}$$

So, 


If both rate constants are affected in the same way (multiplied by the same speed-up factor) so that the quotient doesn't change, then also the equilibrium constant does not change, no shifting, no change in Gibbs free energy.
That would be the case with your catalyst as that facilitates both forward and reverse reaction by lowering the energy of the transition state in the elementary reaction leading to just such a multiplicative increase in reaction rate - but no effect on the energy of educts or products (thus, no change in $\Delta G$).

This has an interesting link to the impossibility of constructing a "chemical" perpetuum mobile by use of a catalyst. 
If the rate constant of either forward or reverse reaction (or both, but not by the same factor) change, the equilibrium shifts.
But, yes, that also means that Gibbs free energy has changed.
Thus, there must be a change in free energy of educts and/or products as the energy of the transition state does not affect $\Delta G$, it only affects the reaction rates.
Whether this is possible hinges on the question whether we consider it still the same reaction $\ce{A + B &lt;=&gt; C + D}$ or whether we say that now we have another equilibrium, say, $\ce{A' + B &lt;=&gt; C + D}$ or $\ce{A' + B' &lt;=&gt; C' + D'}$ which of course can have different equilibrium constants. An example where I think one may argue either way would be the dependency of equilibrium constants on solvent.
(And, btw, we still cannot build a perpetuum mobile: solvation energy makes the difference)