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Edit. I understood OP's question as follows: If $G$ acts by isometries on a geodesic metric space $X$, and if $Y$ is a geodesic metric space quasi-isometric to $X$, does it imply that $G$ acts non-trivially on $Y$?

Some general results:

• Any hyperbolic group is quasi-isometric to a CAT(0) cube complex, so if you take any hyperbolic group $G$ satisfying Kazhdan's property (T), for instance a cocompact lattice in the quaternionic hyperbolic space or a random group (taking a good density), then it acts geometrically on a space which is quasi-isometric to a CAT(0) cube complex (namely itself) but any isometric action of $G$ on a CAT(0) cube complex has a fixed point.
• Any acylindrically hyperbolic group admits a non-trivial (more precisely, acylindrical) action on a quasi-tree. However, plenty of acylindrically hyperbolic groups satisfy Serre's property (FA), so that they cannot act on a tree without fixing a point.

For a more elementary example, consider the triangle group $$T= \langle a,b,c \mid a^2=b^2=c^2=(ab)^3=(ac)^3=(bc)^3=1 \rangle.$$ It is the symmetry group of the tesselation of the plane by equilateral triangles. Such a triangle complex is quasi-isometric to the square complex $X$ obtained by tiling the plane with squares. However, as noticed in one of my previous answers, $\mathrm{Isom}(X)= (D_\infty \times D_\infty) \rtimes \mathbb{Z}_2$ does not contain ordre-three elements, so that any action of $T$ on $X$ must fix a point.

Edit. I understood OP's question as follows: If $G$ acts "nicely" by isometries on a geodesic metric space $X$, and if $Y$ is a geodesic metric space quasi-isometric to $X$, does it imply that $G$ acts non-trivially on $Y$?

Some general results:

• Any hyperbolic group is quasi-isometric to a CAT(0) cube complex, so if you take any hyperbolic group $G$ satisfying Kazhdan's property (T), for instance a cocompact lattice in the quaternionic hyperbolic space or a random group (taking a good density), then it acts geometrically on a space which is quasi-isometric to a CAT(0) cube complex (namely itself) but any isometric action of $G$ on a CAT(0) cube complex has a fixed point.
• Any acylindrically hyperbolic group admits a non-trivial (more precisely, acylindrical) action on a quasi-tree. However, plenty of acylindrically hyperbolic groups satisfy Serre's property (FA), so that they cannot act on a tree without fixing a point.

For a more elementary example, consider the triangle group $$T= \langle a,b,c \mid a^2=b^2=c^2=(ab)^3=(ac)^3=(bc)^3=1 \rangle.$$ It is the symmetry group of the tesselation of the plane by equilateral triangles. Such a triangle complex is quasi-isometric to the square complex $X$ obtained by tiling the plane with squares. However, as noticed in one of my previous answers, $\mathrm{Isom}(X)= (D_\infty \times D_\infty) \rtimes \mathbb{Z}_2$ does not contain ordre-three elements, so that any action of $T$ on $X$ must fix a point.

Edit. I understood OP's question as follows: If $G$ acts by isometries on a geodesic metric space $X$, and if $Y$ is a geodesic metric space quasi-isometric to $X$, does it imply that $G$ acts non-trivially on $Y$?

Some general results:

• Any hyperbolic group is quasi-isometric to a CAT(0) cube complex, so if you take any hyperbolic group $G$ satisfying Kazhdan's property (T), for instance a cocompact lattice in the quaternionic hyperbolic space or a random group (taking a good density), then it acts geometrically on a space which is quasi-isometric to a CAT(0) cube complex (namely itself) but any isometric action of $G$ on a CAT(0) cube complex as a fixed point.
• Any acylindrically hyperbolic group admits a non-trivial (more precisely, acylindrical) action on a quasi-tree. However, plenty of acylindrically hyperbolic groups satisfying Serre's property (FA), so that they cannot act on a tree without fixing a point.

For a more elementary example, consider the triangle group $$T= \langle a,b,c \mid a^2=b^2=c^2=(ab)^3=(ac)^3=(bc)^3=1 \rangle.$$ It is the symmetry group of the tesselation of the plane by equilateral triangles. Such a triangle complex is quasi-isometric to the square complex $X$ obtained by tiling the plane with squares. However, as noticed in one of my previous answers, $\mathrm{Isom}(X)= (D_\infty \times D_\infty) \rtimes \mathbb{Z}_2$ does not contain ordre-three elements, so that any action of $T$ on $X$ must fix a point.

Edit. I understood OP's question as follows: If $G$ acts by isometries on a geodesic metric space $X$, and if $Y$ is a geodesic metric space quasi-isometric to $X$, does it imply that $G$ acts non-trivially on $Y$?

Some general results:

• Any hyperbolic group is quasi-isometric to a CAT(0) cube complex, so if you take any hyperbolic group $G$ satisfying Kazhdan's property (T), for instance a cocompact lattice in the quaternionic hyperbolic space or a random group (taking a good density), then it acts geometrically on a space which is quasi-isometric to a CAT(0) cube complex (namely itself) but any isometric action of $G$ on a CAT(0) cube complex has a fixed point.
• Any acylindrically hyperbolic group admits a non-trivial (more precisely, acylindrical) action on a quasi-tree. However, plenty of acylindrically hyperbolic groups satisfy Serre's property (FA), so that they cannot act on a tree without fixing a point.

For a more elementary example, consider the triangle group $$T= \langle a,b,c \mid a^2=b^2=c^2=(ab)^3=(ac)^3=(bc)^3=1 \rangle.$$ It is the symmetry group of the tesselation of the plane by equilateral triangles. Such a triangle complex is quasi-isometric to the square complex $X$ obtained by tiling the plane with squares. However, as noticed in one of my previous answers, $\mathrm{Isom}(X)= (D_\infty \times D_\infty) \rtimes \mathbb{Z}_2$ does not contain ordre-three elements, so that any action of $T$ on $X$ must fix a point.

Some general results:

• Any hyperbolic group is quasi-isometric to a CAT(0) cube complex, so if you take any hyperbolic group $G$ satisfying Kazhdan's property (T), for instance a cocompact lattice in the quaternionic hyperbolic space or a random group (taking a good density), then it acts geometrically on a space which is quasi-isometric to a CAT(0) cube complex (namely itself) but any isometric action of $G$ on a CAT(0) cube complex as a fixed point.
• Any acylindrically hyperbolic group admits a non-trivial (more precisely, acylindrical) action on a quasi-tree. However, plenty of acylindrically hyperbolic groups satisfying Serre's property (FA), so that they cannot act on a tree without fixing a point.

For a more elementary example, consider the triangle group $$T= \langle a,b,c \mid a^2=b^2=c^2=(ab)^3=(ac)^3=(bc)^3=1 \rangle.$$ It is the symmetry group of the tesselation of the plane by equilateral triangles. Such a triangle complex is quasi-isometric to the square complex $X$ obtained by tiling the plane with squares. However, as noticed in one of my previous answers, $\mathrm{Isom}(X)= (D_\infty \times D_\infty) \rtimes \mathbb{Z}_2$ does not contain ordre-three elements, so that any action of $T$ on $X$ must fix a point.

Edit. I understood OP's question as follows: If $G$ acts by isometries on a geodesic metric space $X$, and if $Y$ is a geodesic metric space quasi-isometric to $X$, does it imply that $G$ acts non-trivially on $Y$?

Some general results:

• Any hyperbolic group is quasi-isometric to a CAT(0) cube complex, so if you take any hyperbolic group $G$ satisfying Kazhdan's property (T), for instance a cocompact lattice in the quaternionic hyperbolic space or a random group (taking a good density), then it acts geometrically on a space which is quasi-isometric to a CAT(0) cube complex (namely itself) but any isometric action of $G$ on a CAT(0) cube complex as a fixed point.
• Any acylindrically hyperbolic group admits a non-trivial (more precisely, acylindrical) action on a quasi-tree. However, plenty of acylindrically hyperbolic groups satisfying Serre's property (FA), so that they cannot act on a tree without fixing a point.

For a more elementary example, consider the triangle group $$T= \langle a,b,c \mid a^2=b^2=c^2=(ab)^3=(ac)^3=(bc)^3=1 \rangle.$$ It is the symmetry group of the tesselation of the plane by equilateral triangles. Such a triangle complex is quasi-isometric to the square complex $X$ obtained by tiling the plane with squares. However, as noticed in one of my previous answers, $\mathrm{Isom}(X)= (D_\infty \times D_\infty) \rtimes \mathbb{Z}_2$ does not contain ordre-three elements, so that any action of $T$ on $X$ must fix a point.