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Suppose that topological group $G$ acting on topological space $X$. If the set $\left\{ \left[ G_{x}\right] :x\in X\right\} $ is finite, where $\left[ G_{x}\right] $ denotes the conjugacy class of the isotropy subgroup $G_{x}$ in $G$, then the group $G$ is said to act on a space $X$ with finitely many orbit types (FMOT). If the set $\left\{ \left[ \left( G_{x}\right) _{0}% \right] :x\in X\right\} $ is finite, where $\left( G_{x}\right) _{0}$ denotes the identity component of the isotropy subgroup $G_{x}$ in $G$, then the group $G$ is said to act on a space $X$ with finitely many connective orbit types (FMCOT).

I asked before that Fixed point set for a subcircle of torus actions if $T$ is a torus and $X$ is a $T$-space with finitely many connective orbit type, then there is a subcircle $L=\mathbb{S}% ^{1}\subset T$ such that $X^{L}=X^{T}$ (fixed point sets of the action). I couldn't get a complete answer to this question.

While I was thinking about this question, I came across something like this: Closed subgroups of torus $T=\mathbb{S}^{1}\times \mathbb{S}^{1}\times \cdots \times \mathbb{S}^{1}$ has type $\mathbb{S}^{1}\times \mathbb{S}% ^{1}\times \cdots \times \mathbb{S}^{1}\times %TCIMACRO{\U{2124} }% %BeginExpansion \mathbb{Z} %EndExpansion _{n_{1}}\times %TCIMACRO{\U{2124} }% %BeginExpansion \mathbb{Z} %EndExpansion _{n_{2}}\times \cdots \times %TCIMACRO{\U{2124} }% %BeginExpansion \mathbb{Z} %EndExpansion _{n_{k}}$. So connected closed subgroups of $T$ has type $\mathbb{S}% ^{1}\times \mathbb{S}^{1}\times \cdots \times \mathbb{S}^{1}\times \left\{ 1\right\} \times \left\{ 1\right\} \times \cdots \times \left\{ 1\right\} $. Isn't there already finite number of connective orbit types for torus actions on any topological space? Where am i doing wrong?

Add: I find this thesis ''Computation of Fixed Point Data from Equivariant Cohomology by Courtney K. Taylor''

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    $\begingroup$ No there are infinitely many non-cconjugate subgroups of the same type. The crucial point here is the difference between inner and non-inner automorphism. For example, for any two embeddings $f_1,f_2:S^1\to T^2$ there is an automorphism $\sigma$ of $T^2$ such that $f_2=\sigma\circ f_1$. However there need not be an inner automorphsm. $\endgroup$
    – HenrikRüping
    Commented Dec 13, 2023 at 11:01
  • $\begingroup$ @HenrikRüping Since $T$ is abelian, the conjugacy class $[(G_x)_0]$ consists of only $(G_x)_0$. So $(G_x)_0$ equals to $\mathbb{S}^{1}\times \mathbb{S}^{1}\times \cdots \times \mathbb{S}^{1}\times \left\{1\right\} \times \left\{ 1\right\} \times \cdots \times \left\{ 1\right\}$? $\endgroup$
    – Mehmet Onat
    Commented Dec 13, 2023 at 11:30
  • $\begingroup$ @HenrikRüping For any $x \in X$, $G_x=\mathbb{S}^{1}\times \mathbb{S}^{1}\times \cdots \times \mathbb{S}^{1}\times \mathbb{Z}_{n_{1}}\times \mathbb{Z}_{n_{2}}\times \cdots \times \mathbb{Z}_{n_{k}}$? $\endgroup$
    – Mehmet Onat
    Commented Dec 13, 2023 at 11:36
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    $\begingroup$ Well maybe we should have a closer look at the example of $T^2$. Then any closed subgroup of $T^2$ is either trivial, the whole of $T^2$ or of the form $\{(at,bt)\mid t\in S^1\}$ for some coprime $a,b\in \mathbb{Z}$. They are all nonconjugate and they are all "of the type" $S^1\times \{1\}$. The map that assigns to a conjugacy class of a closed subgroup its type thus sends infinitely many different conjugacy classes to the same type. $\endgroup$
    – HenrikRüping
    Commented Dec 13, 2023 at 13:04
  • $\begingroup$ @HenrikRüping Thank you for your time and patient reply. $\endgroup$
    – Mehmet Onat
    Commented Dec 13, 2023 at 13:31

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