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Beautiful problem. Note that $P(p,p)$ yields $\mathbf F(p)^2=\mathbf F(p\circ p)$. In particular, $\mathbf F(\text{id})^2=\mathbf F(\text{id})$ so $\mathbf F(\text{id})\in\{0,1\}$. If $\mathbf F(\text{id})=0$, then $P(p,\text{id})$ gives us $\mathbf F(p)^2=\mathbf F(p\circ p)+2\mathbf F(p)$ so $\mathbf F(p)=0$ for all $p\in\mathcal{B}$, which works. Now, let's assume that $\mathbf F(\text{id})=1$. Plugging $\mathbf F(p)^2=\mathbf F(p\circ p)$ in $P(p,q),$ we get \[2\mathbf F(p)\mathbf F(q)=\mathbf F(p\circ q)+\mathbf F(q\circ p)\]which we will call $Q(p,q)$. Note that $Q(p,p^{-1})$ yields $\mathbf F(p)\mathbf F(p^{-1})=1$. By combining $Q(p,p^{-1}\circ q)$ and $Q(p^{-1}, q\circ p)$, it follows that \begin{align*}\mathbf F(q)+\mathbf F(p^{-1}\circ q\circ p) &= 2\mathbf F(p)\mathbf F(p^{-1}\circ q) \\ &= 2\mathbf F(p^{-1})\mathbf F(q\circ p)\end{align*}Therefore, $\mathbf F(p)^2\mathbf F(p^{-1}\circ q)=\mathbf F(q\circ p)$. Letting $q=p\circ q\circ q$ in the latter, we finally get \[\mathbf F(p)^2\mathbf F(q)^2=\mathbf F(p)^2\mathbf F(q\circ q)=\mathbf F(p\circ q\circ q\circ p).\]By switching $p$ and $q$ we also get $\mathbf F(p)^2\mathbf F(q)^2=\mathbf F(q\circ p\circ p\circ q)$. By using these two identities in $Q(p\circ q,q\circ p)$ we have \[2\mathbf F(p\circ q)\mathbf F(q\circ p)=2\mathbf F(p)^2\mathbf F(q)^2\]which, combined with $Q(p,q)$, that is $\mathbf F(p\circ q)+\mathbf F(q\circ p)=2\mathbf F(p)\mathbf F(q)$, ultimately yields \[\mathbf F(p\circ q)=\mathbf F(q\circ p)=\mathbf F(p)\mathbf F(q).\]Knowing that $\mathbf F(p\circ q)=\mathbf F(p)\mathbf F(q)$ for all $p,q\in\mathcal{B},$ we shall now start to compute $\mathbf F$. This is where $\mathbf F(\text{id})=1$ comes into play. Recall that $\mathbf F(p)^2=\mathbf F(p\circ p)$. Therefore, if $p$ is an involution, $|\mathbf F(p)|=1$. Claim. For any involution $p$, we have $\mathbf F(p)=1$. Proof. We firstly show that any involution with infinitely many transpositions satisfies the claim. Let $p$ be such an involution. It suffices to show that $p$ may be expressed as $q\circ q$, since then $\mathbf F(p)=\mathbf F(q\circ q)=\mathbf F(q)^2>0$ so as $|\mathbf F(p)|=1$ it follows that $\mathbf F(p)=1$. Indeed, by letting $q$ be a collection of (adequately chosen) cycles of length four and fixed points, we have $q\circ q=p$, since the composition of $q$ with itself decomposes each cycle of length four into two transpositions and conserves fixed points: \begin{align*}q: a_1\to b_1\to a_2\to b_2\to a_1 &\implies q\circ q:\begin{cases}a_1\to a_2\to a_1 \\ b_1\to b_2\to b_1\end{cases} \\ q:a\to a &\implies q\circ q: a\to a\end{align*}For an involution $p$, let $T_p$ denote the set of transpositions of $p$. We know that the claim holds for all $p$ satisfying $|T_p|=\infty$. Consider an involution $p$ such that $|T_p|$ is finite. Pick another involution $r$ satisfying $T_p\subset T_r$ and $|T_r|=\infty$. Finally, choose an involution $s$ satisfying $T_s=T_r\setminus T_p$. Then, $p=r\circ s$, since the common transpositions of $r$ and $s$ cancel out. Therefore, $\mathbf F(p)=\mathbf F(r)\mathbf F(s)=1$, finishing the proof. $\square$ Lemma. Any bijective function is the composition of two involutions. Proof: Decompose our bijection into cycles of finite length and two-sided infinite sequences, that is, orbits of the form \[\cdots\to n_{-1}\to n_0\to n_1\to\cdots.\]It clearly suffices to prove the lemma for each such component independently. Consider a cycle $n_1\to\cdots\to n_k\to n_1$. We express it as $u\circ v$, where $u:n_i\to n_{-i}$ and $v:n_i\to n_{1-i}$, the indices being taken modulo $k$. Now, consider an infinite sequence $\cdots\to n_{-1}\to n_0\to n_1\to\cdots$. The same construction works, this time dropping the modular constraint on the indices; we express it as $u\circ v$, where $u:n_i\to n_{-i}$ and $v:n_i\to n_{1-i}$. $\square$ It follows from the lemma and the claim that $\mathbf F(p)=1$ for all $p\in\mathcal{B}$.

solved with MathLuis By $q\mapsto id$ we get ${F}^2(p) = F(p \circ p)$, (if $F(id)=0$ then clearly $F$ must be $0$ ) Let $P(p,q)$ denote the assertion that, $$2{F}(p){F}(q) = {F}(p \circ q) + {F}(q \circ p)$$Let $p^{-1}$ denote the inverse of the bijection $p$. By $P(p,p^{-1})$ we get ${F}(p){F}(p^{-1}) = 1$, from $P(p^{-1}, q \circ p)$ and $P(p, p^{-1} \circ q)$ we get $${F}(q)+{F}(p^{-1} \circ q \circ p) = 2F(p)F(p^{-1} \circ q) = 2F(p^{-1})F(q \circ p)$$, and therefore $so F(p)^2F(p^{-1} \circ q) = F (q \circ p)$, now by $q \mapsto p \circ q \circ q$ we get $$F(p)^2F(q)^2=F(p \circ q \circ q \circ p)$$, interchange $p,q$, to get $F(p \circ q \circ q \circ p) = F (q \circ p \circ p \circ q)$, and by $P(p \circ q, q \circ p)$ we get $2F(p \circ q)F(q \circ p) = 2F(p)^2F(q)^2$ and now combining this with $P(p,q)$ we get that $$F(p)F(q) = F(p \circ q) = F(q \circ p)$$, now this completes the manipulation part of the problem, now we explore the combinatorial structure. By the above, we instantly get that $F(p)^2=1$ for all involutions. Claim 1: Every involution $p$ can be expressed as composition of two bijections in $\mathbb{N}$, that is, $p = q \circ q$ for some $q$. Proof: We construct $q$. $q$ preserves the fixed points of $p$, now take any two cycles $a\mapsto b \mapsto a$ and $c \mapsto d \mapsto c$ then q will $a\mapsto b \mapsto c \mapsto d \mapsto a$, and repeat the process, it is clear that there are no inconsistencies. The above claim yields that $F(p)=1$ for all involutions. Now the following claim finishes to give that $F(p) = 1$ for all bijections $p$. Claim 2: Every bijection in $\mathbb{N}$ can be expressed as composition of two involutions. Proof: The proof again exploits the cycle structure of bijections, Consider the orbit of any number $x$ $\cdots \mapsto f^-1(x) \mapsto x \mapsto f(x) \mapsto \cdots$, say $a_1 \mapsto a_2 \cdots \mapsto a_k$ is a cycle, then, it is clear that it suffices to prove the statement for each individual component, Take $p(a_k) = a_{-k}$ and $q(a_k) = a_{1-k}$. $\blacksquare$