Contrary to our intuition the answer is negative: There are smooth actions of compact groups $G\times D^n\to D^n$ (even finite groups) on closed $n$-disk $D^n$ (for $n$ sufficiently large) so that the action is isometric with respect to some Riemannian metric on $D^n$ and has no point in $D^n$ fixed by the group $G$.
By considering the distance function defined by the Riemannian metric, one obtains actions of compact groups of isometries of the resulting metric space, which do not have a fixed point in $D^n$.
Here is where the examples are coming from:
Given a smooth manifold $M$ (possibly with boundary) and a smooth action of a compact group $G$ on $M$, denoted $G\times M\to M$, there exists a $G$-invaraint Riemannian metric $ds^2$ on $M$. This standard fact is proven by taking an arbitrary Riemannian metric $ds_o^2$ on $M$ and averaging it under the action of the group $G$. If $G$ is finite, the averaging procedures is just
$$
ds^2= \frac{1}{|G|} \sum_{g\in G} g^*(ds_o^2).
$$
If $G$ is compact one replaces the sum with the integral over the Haar measure of $G$.
Thus, it remains to find compact groups acting smoothly on $D^n$ without a fixed point. The first examples of this type were constructed by Conner and Richardson, in the case when $G$ is the icosahedral group, i.e., the group $I$ of orientation preserving symmetries of the regular 3-dimensional icosahedron. Later on, Oliver constructed an example of a smooth action of $SO(3)$ on $D^8$ which does not have a fixed point (in particular, the subgroup $I<SO(3)$ does not fix a point in $D^8$). You can find a detailed description of Oliver's example, references and further information in the survey:
M.Davis, A survey of results in higher dimensions, In "The Smith Conjecture",
(editors: J. W. Morgan and H. Bass), Academic Press, New York, 1984,
https://people.math.osu.edu/davis.12/old_papers/survey.pdf
On the positive side, if you equip $D^n$ with a Riemannian metric of nonpositive curvature $ds^2$, so that the boundary is convex, then the isometry group of $(D^n, ds^2)$ does have a fixed point in $D^n$: This is a corollary of Cartan's Fixed Point theorem. You can find a proof of the latter for instance in Petersen's "Riemannian Geometry" book. I think, Kirill in his answer was trying to reproduce the proof of Cartan's theorem, without the nonpositive curvature assumption (which, of course, cannot be done).
I have a feeling that this question was already asked at MSE some time ago, but it's easier to write an answer than to find a duplicate. So, here it is.
Let me assume that the noncompact manifold $M$ contains a noncompact connected component $X$. (I will leave it to you to find a fixed-point-free self-map of a manifold which has infinitely many connected components.)
The basic observation is that every noncompact connected manifold $X$ contains a ray, i.e. a closed subset $A$ homeomorphic to $[0,\infty)$, equivalently, that there is an injective proper continuous map $c: [0,\infty)\to X$. See, for instance, Kajelad's answer here.
In fact, the same is true for a much larger class of topological spaces:
Let $X$ be a metrizable, 2nd countable, locally compact, connected, locally path-connected, noncompact space. Then $X$ contains a ray.
Now, back to our manifold. Given a ray $A\subset X$, and a homeomorphism
$$
c: [0,\infty)\to A,
$$
define the continuous map $f: A\to [1,\infty)$, $f(a)=c^{-1}(a)+1$. Now, use the Tietze extension theorem to extend $f$ to a continuous map
$$
g: M\to (0,\infty).
$$
Lastly, take the composition
$$
h: M\to A\subset M, h= c\circ g.
$$
Clearly, $h$ has no fixed points (since its restriction to $A$ has no fixed points).
You can see from the proof that the existence of fixed-point free maps holds in much greater generality, say, for connected, locally path-connected, metrizable, 2nd countable, locally compact, noncompact spaces.
Edit. From the comments, it's clear that you are actually interested in constructing a diffeomorphism without fixed points on a noncompact connected manifold $M$. Here are the key steps in the construction:
i. Every smooth connected noncompact manifold admits a nonvanishing vector field. This was discussed on MSE and MO many times (my list is admittedly incomplete):
January 2011, November 2011, November 2013, November 2018, June 2021,
The most satisfactory answer (to my taste) is the one from December 2020: It avoids the obstruction theory and the details are easy to fill-in.
ii. Given a nonvanishing vector field $X$ on a manifold $M$, one can multiply $X$ by a positive function $\varphi\in C^\infty(M)$ such that the new vector field $Y=\varphi X$ is complete. My favorite way to do so is by using a complete Riemannian metric $g$ on $M$ and normalizing $X$ to a unit vector field on $M$ as it is done in this answer.
iii. Now, one can take the time-one flow $h=F(\cdot, 1)$ of the vector field $Y$. The only way $h$ can have a fixed-point $x$ in $M$ is when $x$ lies on some periodic trajectory $T$ of $Y$, whose length (with respect to the metric $g$ as above) is of the form $1/k$, $k\in {\mathbb N}$. Since $Y$ is nonvanishing, for each compact $K\subset M$ the infimum of lengths of periodic trajectories of $Y$ is bounded from below by some $\epsilon_K>0$. There exists a positive function $\psi\in C^\infty(M)$ such that for each compact $K\subset M$,
$$
\inf \psi|_K < \epsilon_K.
$$
Then the product $Z=\psi Y$ is still a complete vector field (since it is uniformly bounded from above) and, at the same time, the time-1 flow of $Z$ has no fixed points in $M$.
Lastly: it is unclear to me how to prove a similar results (existence of fixed-point free maps) in the PL and topological category, although, I have no doubt that they still hold.
Best Answer
Let $D$ be open disk with center $0$ of radius $1$ in complex plane. Consider holomorphic automorphism of $D$ given by formula $z \mapsto \frac{z+a}{1+\bar a z}$ with $a\in D$. Does it have fixed point if $a\neq 0$ ? [You must solve $z=\frac{z+a}{1+\bar a z}$]
You can also biholomorphically send $D$ to right half plane $H$ of $\mathbb C$ by $z \mapsto \frac{1+z}{1-z}$ and then apply Chris Eagle comment to $H$.