This is false. For example take $f: [0,\infty)\to [0,\infty)$ given by
$$
f(x)=\begin{cases}0& \text{ if } x=0\\
e^{-1/x}& \text{ if } x>0
\end{cases}$$

Then $\lim_{x\to 0}\frac{f(2x)}{f(x)}=\infty$ so $f$ is not quasisymmetric. But $f$ is continuous and monotone hence gives a homeomorphism onto the image when restricted to $[0,1]$. Since $[0,1]$ is compact $f$ is uniformly continuous on it. Same for $f^{-1}$.

Let me now answer the original question (which the OP erased) which asked for a weaker conclusion which is still false.
The question was the following.

Let $f: X\to Y$ be a bijection such that both $f$ and $f^{-1}$ are uniformly continuous. Suppose $X$ is doubling i.e. there exists a constant $C>$ such any ball of radius $r$ can be covered by at most $C$ balls of radius $r/2$.

Question: Does this imply that $Y$ is also doubling?

The answer is NO.

The answer is yes if $f$ is bi-Lipschitz in which case the statement is trivial but any weaker conditions on the modulus of continuity should not be sufficient.

Here is an example of a metric of the closed $2$-disk which is not doubling.

Take a 2-disk $\bar D=\{x^2+y^2\le 1, z=0\}$ in the $xy$ plane in $\mathbb R^3$ and attach 10 vertical intervals of length 2 at different points of the interior of the disk. The doubling constant of the resulting space $\tilde X_1$ on scale 1 will be about 10.

The space $\tilde X_1$ is no longer a disk but we can make it one by replacing the vertical intervals by very thin "fingers" (graphs of bump functions on tiny disks) of the same height 2 so that the the new space $X_1$ is Hausdorff close to $\tilde X_1$ and is a graph of a function from $\bar D$ to $\mathbb R$.

Now take a very small $\epsilon$, take small disk $D_\epsilon$ in $D$ if radius $\epsilon$ away from the fingers and repeat the procedure on $D_\epsilon$ on scale $\epsilon$ by attaching 100 vertical intervals of length $2\epsilon$. Replace the smaller intervals by fingers. then the doubling constant on scale $\epsilon$ will be 100. Iterate. The limit space $X_\infty$ will be a graph of a function on $\bar D$ but will not be doubling. The projection map $f: X_\infty\to \bar D$ is a homeomorphism. It's 1-Lipschitz. The inverse is uniformly continuous because the spaces involved are compact.