Why does the equilibrium constant change on scaling the stoichiometric coefficients of a reaction?
Have a look at this answer of mine.
There you can find the derivation of the formula that defines the equilibrium constant

\begin{equation}
\log \underbrace{\prod_i [a_{i}]^{\nu_{i}}}_{= \, K} = -\frac{\Delta G^{0}}{RT} \qquad \Rightarrow \qquad \log K = -\frac{\Delta G^{0}}{RT} 
\end{equation}

of a general reaction

\begin{equation}
 \ce{\nu_{1}A + \nu_{2}B + ... + \nu_{$x$}C &lt;=&gt; \nu_{$x+1$}D + \nu_{$x+2$}E + ... +\nu_{$n$}F}
\end{equation}

and it is also shown how this definition relates to the maybe more familiar definition

\begin{equation}
 K_{c} = \prod_i [c_{i}]^{\nu_{i}} \ .
\end{equation}

Now, the reason why the equilibrium constants of the general reaction above and the same reaction multiplied with a constant factor $\alpha$:

\begin{equation}
 \ce{\alpha\nu_{1}A + \alpha\nu_{2}B + ... + \alpha\nu_{$x$}C &lt;=&gt; \alpha\nu_{$x+1$}D + \alpha\nu_{$x+2$}E + ... + \alpha\nu_{$n$}F}
\end{equation}

are not the same but are related to each other via:

\begin{equation}
 K_{\alpha} = K_{1}^{\alpha}
\end{equation}

comes from the definition of $K$.

That there must be a difference is also plausible: The defining formula for $K$ relates it to the standard Gibbs free energy of reaction $\Delta G^{0}$ which in turn is defined to be the amount of Gibbs free energy that is produced per formula conversion (meaning that it is defined for an extend of reaction $\xi = 1 \, \text{mol}$).
So, it is the amount of Gibbs free energy produced by the reaction of $\nu_{1}\, \text{mol}$ of $\text{A}$ and $\nu_{2}\, \text{mol}$ of $\text{B}$ etc. to $\nu_{x+1} \, \text{mol}$ of $\text{D}$ and $\nu_{x+2}\, \text{mol}$ $\text{E}$ etc.
But if you multiply the reaction by $\alpha$ the standard Gibbs free energy of reaction is the Gibbs free energy produced by the reaction of $\alpha\nu_{1}\, \text{mol}$ of $\text{A}$ and $\alpha\nu_{2}\, \text{mol}$ of $\text{B}$ etc. to $\alpha\nu_{x+1} \, \text{mol}$ of $\text{D}$ and $\alpha\nu_{x+2}\, \text{mol}$ $\text{E}$ etc. and so there should be $\alpha$ times as much Gibbs free energy as before:

\begin{equation}
 \Delta G^{0}_{\alpha} = \alpha \Delta G^{0}_{1} \ .
\end{equation}

And this is indeed the case when you have $K_{\alpha} = K_{1}^{\alpha}$ because according to the logarithm laws

\begin{equation}
RT \log K_{1}^{\alpha} = RT \alpha \log K_{1} \ .
\end{equation}