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Let $f$ be a Morse function on closed smooth manifold $M$, (one can also suppose that this is a hyperbolic manifold) G is a group which acts effectively and smooth on $M$, f and G (hyperbolic metric) are equivariant in $M$. There many results about deformations of Morse functions. But is there any way to always pertrubate $f$ in a equivariant way? More precisely, is it true that for every $\epsilon$ there exists $g$ - smooth equivariant Morse function which is $|f-g|_{M} < \epsilon$, and if it is true, can one take such $g$ to be a Morse-Smale function? Any links would be much appreciated.

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    What is a Morse-Smale function? I guess you require $g\ne f$. Is $G$ supposed to be a finite group? Once you assume finiteness of $G$, it is easy to construct nontrivial equivariant Morse perturbations of equivariant Morse functions (perturb your function away from the critical set). – 2017-02-09
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    Yeah, seems nice to think about finite groups first. But I did not get argument of yours. It would be nice for you to write more. Thanks anyways. – 2017-02-09
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    https://en.wikipedia.org/wiki/Morse_homology - a link about Morse - Smale condition. – 2017-02-09
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    I wrote finite since isometry groups of hyperbolic manifolds are finite. You should edit your question to include MS function definition (transverse intersections of stable and unstable manifolds). I will write my argument later. – 2017-02-09

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If your function is equivariant, it must take the same value at all points in an orbit. Now look at $S^2$, with $SO(3)$ acting on it: there's only one orbit, so all equivariant functions are constant. So any equivariant function is not a morse function.

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    Anyways I have a family of manifolds with equivariant Morse function on it. The question is not about a possibility of existence of such a functions. – 2017-02-08
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    I got that. I just wanted to point out that there *are* smooth functions that are not close to any equivariant Morse function, and that satisfy all the hypotheses you gave (aside from the "one can also suppose..." which is NOT something you can suppose from the hypotheses you gave). That was particularly easy to show in the case where there are *no* equivariant Morse functions. You might consider rewriting to make it clear that you're talking about ONLY hyperbolic manifolds on which you know there are smooth equivariant Morse functions. Or you might not... – 2017-02-08
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    @JohnHughes: $S^2$ is not a hyperbolic manifold and this is what OP appears to be interested in. – 2017-02-09
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    I know $S^2$ is not hyperbolic. OP's phrasing seems to (ludicrously) suggest that if a smooth closed manifold $M$ has a Morse function, then it's hyperbolic. If OP had asked the question clearly, I wouldn't have bothered giving this answer. No good deed goes unpunished. :) – 2017-02-09
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    Thank you very much for interesting and simple example anyways. – 2017-02-09