Any Springer isomorphism has the desired property.
(Here I'm working over an algebraically closed field, else one should be more careful with the language)

Indeed, let $P$ be any parabolic subgroup
of $G$. Then there is a cocharacter $\lambda:\mathbf{G}_m \to G$ for which $P = P(\lambda)$
is the parabolic subgroup determined by $\lambda$ -- see [Springer, Linear Alg. Groups Prop. 8.4.5].

Explicitly, $P =$ {$x \in G \mid \operatorname{lim}_{t \to 0} \lambda(t) x \lambda(t)^{-1}$ exists}
and the unipotent radical $U = R_u(P)$ is given by $U = $ {$x \in P \mid \operatorname{lim}_{t \to 0} \lambda(t) x \lambda(t) ^{-1} = 1$}; see [Springer, Linear Algebraic Groups, 3.2.13] for more on these limits.

In particular, it follows that $U$ is the set of all $x$ in the unipotent variety for which $\operatorname{lim}_{t \to 0} \lambda(t) x \lambda(t)^{-1}$ exists and is equal to $1$.

Moreover, $\operatorname{Lie}(P)$ and $\operatorname{Lie}(U)$ have similar descriptions
-- e.g. $\operatorname{Lie}(P)$ consists of all $X \in \operatorname{Lie}(G)$ such
that $\operatorname{lim}_{t \to 0} \operatorname{Ad}(\lambda(t))X$ exists.

As before, one finds that $\operatorname{Lie}(U)$ consists in all the $X$ in the nilpotent
variety for which $\operatorname{lim}_{t \to 0} \operatorname{Ad}(\lambda(t))X$ exists and is equal to $0$.

Since a Springer isomorphism $\phi$ is $G$-equivariant (and maps $0 \mapsto 1$), it follows
from these descriptions that $\phi$ maps $\operatorname{Lie}(U)$ isomorphically onto $U$.

I suspect (hope?!) that something like this argument is given in some paper I've written; maybe the one with Donna that Jim mentioned in his comment?

*EDIT:* Actually I wrote down the required statement in section 4 (remark 10) of "Optimal SL(2)-homomorphisms,"
Comment. Math. Helv. 80 (2005), no. 2, 391–426.

`$\phi$`

of the type you consider. For the classical root systems there are explicit maps available (Cayley, etc.); have you looked at these cases? It's also good to add a reference, such as McNinch-Testerman, J. Pure Appl. Algebra 213 (2009), 1346–1363. $\endgroup$ – Jim Humphreys Dec 22 '11 at 22:28