Intuition for why catalyst affects both forward and reverse reactions equally?
I don't know if this qualifies as an intuition answer or not, but it's very easy to show mathematically. The key is that the rate constant is not linearly proportional to the activation energy. By the Arrhenius equation, $k = Ae^{-E_\mathrm{A}/(RT)}$, so the rate constant is dependent on an exponential of the activation energy. The equilibrium constant $K$ is a function of the ratio of the forward and reverse rate constants. For a simple reaction $\ce{A &lt;=&gt;[k_1][k_{-1}] B}$, the equilibrium constant $K=\frac{k_1}{k_{-1}}$.

If we assign the activation energy $E_\mathrm{A,fwd}$ to the forward reaction and $E_\mathrm{A,rev}$ to the reverse, we have 

$$k_1 = Ae^{-E_\mathrm{A,fwd}/(RT)} \quad \text{and} \quad k_{-1} = Ae^{-E_\mathrm{A,rev}/(RT)},$$

so

$$K=\frac{Ae^{-E_\mathrm{A,fwd}/(RT)}}{Ae^{-E_\mathrm{A,rev}/(RT)}}.$$

Using simple rules of exponents, this reduces to

$$K = e^{E_\mathrm{A,rev} - E_\mathrm{A,fwd}}.$$

Now if we subtract a quantity $E_\mathrm{cat}$ from both activation energies, the same derivation gives

$$K = e^{E_\mathrm{A,rev} - E_\mathrm{cat} - (E_\mathrm{A,fwd} - E_\mathrm{cat})} = e^{E_\mathrm{A,rev} - E_\mathrm{A,fwd}},$$

so $K$ is unchanged. 

Intuitively, it may help to think about the fact that the reverse reaction is necessarily composed of exactly the same microscopic states as the forward reaction, just happening in reverse order (by the principle of microscopic reversibility). The catalyst can stabilize one of more of the states on that reaction pathway, but cannot influence what direction that stabilized state reacts in. So making a microscopic state more easily achievable in the forward direction necessarily makes it more easily achievable in the reverse direction.