Le Châtelier's Principle and heat
Before to show you that what you said is false (I'm sorry ^^) be sure you understand the constant of a reaction depends on the temperature.



Let the same reaction you want $$\mathrm{A \rightleftharpoons B}$$

With a constant $\mathrm{K^{\circ}}$. Imagine you heat your system a little with $\mathrm{d}T&gt;0$, then by Van't Hoff's law we get:

$$\frac{\mathrm{d}\ln(K^{\circ})}{\mathrm{d}T}=\frac{\Delta_rH^{\circ}}{RT^2}$$
Where $R$ is the perfect gas constant which is positive. Then $RT^2&gt;0$.
 So $\mathrm{d}\ln(K^{\circ})$ as the same sign as $\Delta_rH^{\circ}\cdot \mathrm{d}T$. 

So if you have an endothermic reaction $\Delta_rH^{\circ}&gt;0$ because $\mathrm{d}T&gt;0$ then, $$\mathrm{d}\ln(K^{\circ})&gt;0$$

Then $K^{\circ}$ will increase with the temperature. If you take $\mathrm{d}T&lt;0$ for an endothermic reaction then the constant of the reaction will decrease with the temperature. 

I let you do the same final reasoning for an exothermic reaction. So the "Le Châtelier principle" is still true. 

Note: If you know the chemical affinity, you can do the same proof, just a bit longer!