Interpolation between iterations of exponentials (and logarithms) I'm interested in finding a continuous and ideally smooth family of real-valued monotonic functions $H_t(x)$ that interpolate between iterations of exponentials, and also iterations of logarithms. So the ideal such family would have the properties


*

*$H_0(x) = x$

*$H_{t+1}(x) = \exp\left(H_t(x)\right)$

*$H_{t-1}(x) = \log\left( H_t(x) \right)$
for all $t$. Do such families exist? If there are more than one, is there a categorization of the collection, or any particular such family considered canonical?
It's clear that we only need really define $H_t$ for $t \in [0, 1]$ in such a way so as to ensure smoothness, so it seems like this should reduce to the question of looking for a family of compositional $n^\textrm{th}$ roots for $\exp$, and seeing if the analytic completion of that family is smooth. 
I'm aware of some similar results, e.g. there being many compositional square roots of the exponential function in $\mathbb{R}$. But from the wikipedia article, it isn't immediately clear to me if this process yields $n^\textrm{th}$ roots of $\exp$, or if so if the set of all $n^\textrm{th}$ roots are unique, or if the induced family 
$$\{H_{q}(x) \; : \; q \in \mathbb{Q} \cap [0, 1] \}$$
can be completed to yield a smooth family.

EDIT: Actually I think we want the following slightly stronger set of properties.


*

*$H_0(x) = x$

*$H_1(x) = \exp(x)$

*$H_\alpha(x) \circ H_\beta(x) = H_{\alpha + \beta}(x)$
 A: If you have Tet(x), then 
$$H_n(x)=\text{Tet}(\text{Tet}^{−1}(x)+n)$$
You could visit https://math.eretrandre.org/tetrationforum/index for questions and details. Also, math.eretrandre.org/tetrationforum/showthread.php?tid=1017 for an implmentation of $\text{Tet}^{−1}(x)$ and $\text{Tet}(x)$. 
Kneser's 1949 paper used exactly that equation to generate the half iterate  of the exponential which the Op would call $H_{0.5}(x)$ from analytic tetration which Kneser showed could be generated from a Riemann mapping.
$$\phi(x)=\exp^{[0.5]}(x);\;\;\;\phi(\phi(x))=\exp(x)$$.
A: The problem isn’t finding a solution but the solution.  That is, finding natural assumptions that determines a unique solution.
Kneser’s solution involves choosing one complex fixed point among the many of exp(x).  Ultimately that choice is arbitrary.  (It might be unique given certain assumptions, but that only pushes the arbitrariness back to those assumptions.)
It’s a hard problem.  See
    “The Fourth Operation: iteration from a purely real perspective”
    http://ariwatch.com/VS/Algorithms/TheFourthOperation.htm
Note the “Afterward” at the end which argues that it is a mistake to view the familiar operations as the beginning of a sequence of operations.
