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I answer here positively the second question (it's completely independent of the first one so it could have been 2 distinct posts).

Let $p\le n$ be prime. Let $C\subset S_n$ be the group generated by the cycle $(1\dots p)$. Let $C$ act on $S_n$ by conjugation. It preserves the set $Q$ of squares. Let $Q^C$ be the $C$-fixed points (that is, the set of squares centralizing $C$). The $C$-action on $Q\smallsetminus Q^C$ is free, hence $Q\smallsetminus Q^C$ has cardinality divisible by $p$. The centralizer $Z_C$ of $C$ is $C\times S_{n-p}$ (where $S_{n-p}$ is the pointwise stabilizer of $\{1,\dots,p\}$). Thus $Q^C=Z_C\cap Q$ by definition. If $Q(Z_C)$ is the set of squares of $Z_C$, then clearly $Q(Z_C)\subset Q^C$.

Actually if $p$ is odd, this is an equality (as noticed by "GH from MO", this equality is not a tautology and has to be checked). Indeed, if we have an element in $Q^C=Z_C\cap Q$, its number of $k$-cycles when $k$ is even is the same as its restriction to $\{p+1,\dots,n\}$ (since the possible $p$-cycle does not matter). So $Q^C=C\times\mathrm{Sq}(S_{n-p})$. Hence $\#(Q^C)=p\mathrm{sq}(n-p)$. So $Q$ has cardinality divisible by $p$.

If $p=2$ the previous argument fails at two points, since then $\mathrm{Sq}(Z_C)=\{1\}\times\mathrm{Sq}(S_{n-2})$ (instead of $C\times\mathrm{Sq}(S_{n-2})$), and since the cycle counting argument falls apart.

So we assume $n\ge 4$ and now redefine $C$ as being generated by the 4-cycle $(1\dots 2^k)$ with $k$ maximal ($k=\lfloor\log_2 n\rfloor$). Then $Q\smallsetminus Q^C$ has even cardinality, and the centralizer of $C$ is $C^2\times S_{n-4}$. We have $Q(Z_C)=C^2\times\mathrm{Sq}(S_{n-4})$ (here $C^2$ means the set of squares in $C$, which is a subgroup of order 2). Also $Q(Z_C)\subset Q^C$ and we need to check that it is an equality.

If some element $w$ belongs to $Q^C=Q\cap Z_C$ and $m$ is even, then its number $n_m=n_m(w)$ of $m$-cycles is even by assumption. Write $w=uv$ according to the decomposition $Z_C=C\times S_{n-2^k}$. If $m$ is not a power of $2$, then $u$ has no $m$-cycle, so $n_m(v)=n_m$. If $m$ is a power of 2 and $m<2^k$, then the number of $m$-cycles in $u$ is even (equal to either $2^k/m$ or 0), so $n_m(v)$ is even as well. If $m\ge 2^k$ we have a contradiction because $n-2^k<2^k$, so the only possibility is that $m=2^k$ and the $m$-cycle is supported by $\{1,\dots,2^k\}$, but then $n_m(w)=1$ and $w$ is not a square. Thus $u$ and $v$ are both squares in $C$ and $S_{n-2^k}$ respectively. So $Q^C=C^2\times\mathrm{Sq}(S_{n-4})$, which has even cardinal and thus $Q$ also has even cardinal. [Edit: I modified by initial argument for $p=2$ in which I supposed $k=2$ instead of $k$ maximal, but then $Q(Z_C)\subset Q^C$ can be strict as we see if $n\ge 8$ by taking the element $(1234)(5678)$. So the point noticed by "GH from MO" is not insignificant.]

I answer here positively the second question (it's completely independent of the first one so it could have been 2 distinct posts).

Let $p\le n$ be prime. Let $C\subset S_n$ be the group generated by the cycle $(1\dots p)$. Let $C$ act on $S_n$ by conjugation. It preserves the set $Q$ of squares. Let $Q^C$ be the $C$-fixed points (that is, the set of squares centralizing $C$). The $C$-action on $Q\smallsetminus Q^C$ is free, hence $Q\smallsetminus Q^C$ has cardinality divisible by $p$. The centralizer $Z_C$ of $C$ is $C\times S_{n-p}$ (where $S_{n-p}$ is the pointwise stabilizer of $\{1,\dots,p\}$). Thus $Q^C=Z_C\cap Q$ by definition. If $Q(Z_C)$ is the set of squares of $Z_C$, then clearly $Q(Z_C)\subset Q^C$.

Actually if $p$ is odd, this is an equality (as noticed by "GH from MO", this equality is not a tautology and has to be checked). Indeed, if we have an element in $Q^C=Z_C\cap Q$, its number of $k$-cycles when $k$ is even is the same as its restriction to $\{p+1,\dots,n\}$ (since the possible $p$-cycle does not matter). So $Q^C=C\times\mathrm{Sq}(S_{n-p})$. Hence $\#(Q^C)=p\mathrm{sq}(n-p)$. So $Q$ has cardinality divisible by $p$.

If $p=2$ the previous argument fails at two points, since then $\mathrm{Sq}(Z_C)=\{1\}\times\mathrm{Sq}(S_{n-2})$ (instead of $C\times\mathrm{Sq}(S_{n-2})$), and since the cycle counting argument falls apart.

So we assume $n\ge 4$ and now redefine $C$ as being generated by the 4-cycle $(1\dots 2^k)$ with $k$ maximal ($k=\lfloor\log_2 n\rfloor$). Then $Q\smallsetminus Q^C$ has even cardinality, and the centralizer of $C$ is $C^2\times S_{n-4}$. We have $Q(Z_C)=C^2\times\mathrm{Sq}(S_{n-4})$ (here $C^2$ means the set of squares in $C$, which is a subgroup of order 2). Also $Q(Z_C)\subset Q^C$ and we need to check that it is an equality.

If some element $w$ belongs to $Q^C=Q\cap Z_C$ and $m$ is even, then its number $n_m=n_m(w)$ of $m$-cycles is even by assumption. Write $w=uv$ according to the decomposition $Z_C=C\times S_{n-2^k}$. If $m$ is not a power of $2$, then $u$ has no $m$-cycle, so $n_m(v)=n_m$. If $m$ is a power of 2 and $m<2^k$, then the number of $m$-cycles in $u$ is even (equal to either $2^k/m$ or 0), so $n_m(v)$ is even as well. If $m\ge 2^k$ we have a contradiction because $n-2^k<2^k$, so the only possibility is that $m=2^k$ and the $m$-cycle is supported by $\{1,\dots,2^k\}$, but then $n_m(w)=1$ and $w$ is not a square. Thus $u$ and $v$ are both squares in $C$ and $S_{n-2^k}$ respectively. So $Q^C=C^2\times\mathrm{Sq}(S_{n-4})$, which has even cardinality and thus $Q$ also has even cardinality. [Edit: I modified by initial argument for $p=2$ in which I supposed $k=2$ instead of $k$ maximal, but then $Q(Z_C)\subset Q^C$ can be strict as we see if $n\ge 8$ by taking the element $(1234)(5678)$. So the point noticed by "GH from MO" is not insignificant.]

I answer here positively the second question (it's completely independent of the first one so it could have been 2 distinct posts).

Let $p\le n$ be prime. Let $C\subset S_n$ be the group generated by the cycle $(1\dots p)$. Let $C$ act on $S_n$ by conjugation. It preserves the set $Q$ of squares. Let $Q^C$ be the $C$-fixed points (that is, the set of squares centralizing $C$). The $C$-action on $Q\smallsetminus Q^C$ is free, hence $Q\smallsetminus Q^C$ has cardinality divisible by $p$. The centralizer of $C$ is $C\times S_{n-p}$ (where $S_{n-p}$ is the pointwise stabilizer of $\{1,\dots,p\}$), so if $p$ is odd its number of squares is $\#(Q^C)=p\mathrm{sq}(n-p)$. So $Q$ has cardinality divisible by $p$.

If $p=2$ the previous argument fails since then $\#(Q^C)=\mathrm{sq}(n-2)$ (instead of $2\mathrm{sq}(n-2)$). So we assume $n\ge 4$ and now let $C$ be generated by the 4-cycle $(1234)$. Then $Q\smallsetminus Q^C$ has even cardinality, and the centralizer of $C$ is $C^2\times S_{n-4}$, and has $\#(Q^C)=2\mathrm{sq}(n-4)$ squares (here $C^2$ means the set of squares in $C$, which is a subgroup of order 2). Thus $Q$ has even cardinality.

I answer here positively the second question (it's completely independent of the first one so it could have been 2 distinct posts).

Let $p\le n$ be prime. Let $C\subset S_n$ be the group generated by the cycle $(1\dots p)$. Let $C$ act on $S_n$ by conjugation. It preserves the set $Q$ of squares. Let $Q^C$ be the $C$-fixed points (that is, the set of squares centralizing $C$). The $C$-action on $Q\smallsetminus Q^C$ is free, hence $Q\smallsetminus Q^C$ has cardinality divisible by $p$. The centralizer $Z_C$ of $C$ is $C\times S_{n-p}$ (where $S_{n-p}$ is the pointwise stabilizer of $\{1,\dots,p\}$). Thus $Q^C=Z_C\cap Q$ by definition. If $Q(Z_C)$ is the set of squares of $Z_C$, then clearly $Q(Z_C)\subset Q^C$.

Actually if $p$ is odd, this is an equality (as noticed by "GH from MO", this equality is not a tautology and has to be checked). Indeed, if we have an element in $Q^C=Z_C\cap Q$, its number of $k$-cycles when $k$ is even is the same as its restriction to $\{p+1,\dots,n\}$ (since the possible $p$-cycle does not matter). So $Q^C=C\times\mathrm{Sq}(S_{n-p})$. Hence $\#(Q^C)=p\mathrm{sq}(n-p)$. So $Q$ has cardinality divisible by $p$.

If $p=2$ the previous argument fails at two points, since then $\mathrm{Sq}(Z_C)=\{1\}\times\mathrm{Sq}(S_{n-2})$ (instead of $C\times\mathrm{Sq}(S_{n-2})$), and since the cycle counting argument falls apart. 

So we assume $n\ge 4$ and now redefine $C$ as being generated by the 4-cycle $(1\dots 2^k)$ with $k$ maximal ($k=\lfloor\log_2 n\rfloor$). Then $Q\smallsetminus Q^C$ has even cardinality, and the centralizer of $C$ is $C^2\times S_{n-4}$. We have $Q(Z_C)=C^2\times\mathrm{Sq}(S_{n-4})$ (here $C^2$ means the set of squares in $C$, which is a subgroup of order 2). Also $Q(Z_C)\subset Q^C$ and we need to check that it is an equality.

If some element $w$ belongs to $Q^C=Q\cap Z_C$ and $m$ is even, then its number $n_m=n_m(w)$ of $m$-cycles is even by assumption. Write $w=uv$ according to the decomposition $Z_C=C\times S_{n-2^k}$. If $m$ is not a power of $2$, then $u$ has no $m$-cycle, so $n_m(v)=n_m$. If $m$ is a power of 2 and $m<2^k$, then the number of $m$-cycles in $u$ is even (equal to either $2^k/m$ or 0), so $n_m(v)$ is even as well. If $m\ge 2^k$ we have a contradiction because $n-2^k<2^k$, so the only possibility is that $m=2^k$ and the $m$-cycle is supported by $\{1,\dots,2^k\}$, but then $n_m(w)=1$ and $w$ is not a square. Thus $u$ and $v$ are both squares in $C$ and $S_{n-2^k}$ respectively. So $Q^C=C^2\times\mathrm{Sq}(S_{n-4})$, which has even cardinal and thus $Q$ also has even cardinal. [Edit: I modified by initial argument for $p=2$ in which I supposed $k=2$ instead of $k$ maximal, but then $Q(Z_C)\subset Q^C$ can be strict as we see if $n\ge 8$ by taking the element $(1234)(5678)$. So the point noticed by "GH from MO" is not insignificant.]

I answer here positively the second question (it's completely independent of the first one so it could have been 2 distinct posts).

Let $p\le n$ be prime. Let $C\subset S_n$ be the group generated by the cycle $(1\dots p)$. Let $C$ act on $S_n$ by conjugation. It preserves the set $Q$ of squares. Let $Q^C$ be the $C$-fixed points (that is, the set of squares centralizing $C$). The $C$-action on $Q\smallsetminus Q^C$ is free, hence $Q\smallsetminus Q^C$ has cardinality divisible by $p$. The centralizer of $C$ is $C\times S_{n-p}$ (where $S_{n-p}$ is the pointwise stabilizer of $\{1,\dots,p\}$), so if $p$ is odd its number of squares is $\#(C^Q)=p\mathrm{sq}(n-p)$. So $Q$ has cardinality divisible by $p$.

If $p=2$ the previous argument fails since then $\#(C^Q)=\mathrm{sq}(n-2)$ (instead of $2\mathrm{sq}(n-2)$). So we assume $n\ge 4$ and now let $C$ be generated by the 4-cycle $(1234)$. Then $Q\smallsetminus Q^C$ has even cardinality, and the centralizer of $C$ is $C^2\times S_{n-4}$, and has $\#(Q^C)=2\mathrm{sq}(n-4)$ squares (here $C^2$ means the set of squares in $C$, which is a subgroup of order 2). Thus $Q$ has even cardinality.

I answer here positively the second question (it's completely independent of the first one so it could have been 2 distinct posts).

Let $p\le n$ be prime. Let $C\subset S_n$ be the group generated by the cycle $(1\dots p)$. Let $C$ act on $S_n$ by conjugation. It preserves the set $Q$ of squares. Let $Q^C$ be the $C$-fixed points (that is, the set of squares centralizing $C$). The $C$-action on $Q\smallsetminus Q^C$ is free, hence $Q\smallsetminus Q^C$ has cardinality divisible by $p$. The centralizer of $C$ is $C\times S_{n-p}$ (where $S_{n-p}$ is the pointwise stabilizer of $\{1,\dots,p\}$), so if $p$ is odd its number of squares is $\#(Q^C)=p\mathrm{sq}(n-p)$. So $Q$ has cardinality divisible by $p$.

If $p=2$ the previous argument fails since then $\#(Q^C)=\mathrm{sq}(n-2)$ (instead of $2\mathrm{sq}(n-2)$). So we assume $n\ge 4$ and now let $C$ be generated by the 4-cycle $(1234)$. Then $Q\smallsetminus Q^C$ has even cardinality, and the centralizer of $C$ is $C^2\times S_{n-4}$, and has $\#(Q^C)=2\mathrm{sq}(n-4)$ squares (here $C^2$ means the set of squares in $C$, which is a subgroup of order 2). Thus $Q$ has even cardinality.