Second-step activation energy in SN1 reactions
The $\mathrm{S_N1}$ reaction has two steps i) the formation of carbocation and 2) the formation of product.

(image taken from libretexts)
In the second step an unstable carbocation reacts with the nucleophile. You are asking why this step has to have an activation energy. You can rationalize that by considering the principle of microscopic reversibility. Think about what would happen if the $\mathrm{S_N1}$ reaction went in reverse—first the nucleophile leaves the product, forming a carbocation, and then the carbocation joins with the leaving group to give the reactant. Obviously when the nucleophile leaves the prdouct, and carbocation forms, there is an activation barrier for that.
So regardless of the themodynamics of the total reaction, whenever you detach a functional group and form a
(carbocation + anion), there has to be an activation barrier i.e. a transition state that is higher in energy than both (carbocation + anion) and the initial state. Which means that if you wish to reattach an anion to the carbocation, there will also be an activation energy.