Iwasawa decomposition of inverse Let $G$ be a semisimple rank one Lie group with finite center. Let $G=KAN$ be the Iwasawa decomposition with $\mathfrak{a}=$Lie($A)=\text{span}\ H$. Then if  $G\ni g=kan, a=exp(tH)$ is it true that $$g^{-1}=\tilde k exp(-tH) \tilde n$$ is the decomposition of $g^{-1}$ where $\tilde k\in K, \tilde n\in N$?

The answer is no. Consider $G=SL(2,\mathbb{R})$. Let
$$
g=kan=
\left(\begin{matrix}
0 & -1 \\
1 & 0
\end{matrix}\right) \cdot
\left(\begin{matrix}
1 & 0 \\
0 & 1
\end{matrix}\right) \cdot
\left(\begin{matrix}
1 & 1 \\
0 & 1
\end{matrix}\right) =
\left(\begin{matrix}
0 & -1 \\
1 & 1
\end{matrix}\right).
$$
We can write instead
$$
g=n'a'k'= \left(\begin{matrix}
1 & -\frac{1}{2} \\
0 & 1
\end{matrix}\right) \cdot 
\left(\begin{matrix}
\frac{1}{\sqrt{2}} & 0 \\
0& \sqrt{2}
\end{matrix}\right) \cdot 
\left(\begin{matrix}
\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}
\end{matrix}\right) = 
\left(\begin{matrix}
1 & 1 \\
0 & 1
\end{matrix}\right).
$$
So we get the Iwasawa decomposition
$$
g^{-1}=k'^{-1}a'^{-1}n'^{-1}
$$
for the inverse, but we see
$$
\left(\begin{matrix}
1 & 0 \\
0 & 1
\end{matrix}\right) 
= \exp (-tH) =
a^{-1} \neq a'^{-1}=
\left(\begin{matrix}
\sqrt{2} & 0 \\
0 & \frac{1}{\sqrt{2}}
\end{matrix}\right) 
.
$$
so your suggestion does not hold.
It would be interesting to learn how the $KAN$ and $NAK$ Iwasawa decompositions (and especially their $A$-component) relate to each other.
