Exponential of adjoint map corresponds to conjugation by group element for Lazard Lie ring

Statement with left action convention
Suppose $$L$$ is a Lazard Lie ring, $$G$$ is its Lazard Lie group, and $$\exp:L \to G$$ is the exponential map establishing the bijection between the elements of $$L$$ and the elements of $$G$$. Suppose $$u \in L$$ and $$\operatorname{ad} u$$ denotes the adjoint action by $$u$$, i.e., the map $$x \mapsto [u,x]$$. In particular, it is an inner derivation.

Then, $$\operatorname{ad} u$$ is an exponentiable derivation. Further, if $$\sigma = \exp(\operatorname{ad} u)$$ (note that this is the exponential of a map and is not the same as the Lazard correspondence exponential), then for $$x \in L$$:

$$\exp(\sigma(x)) = \exp(u)\exp(x)\exp(u)^{-1}$$

In other words, under the Lazard correspondence, the exponential of the adjoint map ($$\exp(\operatorname{ad} u)$$) corresponds to the conjugation by the group element $$\exp(u)$$.

Related facts

 * Exponential map commutes with adjoint action is the version for real Lie groups
 * Matrix exponential commutes with conjugation is the version for linear Lie groups
 * Exponential of derivation is automorphism under suitable nilpotency assumptions