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Not really an answer, but maybe slightly more than a comment:

You want to embed the discrete dynamical system $f$ in a flow $\phi_t$ (so $\phi_1=f$), and the most straightforward way to attempt this is to just write down the ODE $\dot{x}=g(x)$ that produces $\phi_t$ and see what conditions $g$ must satisfy. If $H'=1/g$, then the solution is (implicitly) given by $H(\phi_t(x))-H(x)=t$. So in our situation $H$ must satisfy the functional equation $$ H\circ f - H = 1 \quad\quad\quad\quad (F) $$ In principle at least, we can easily describe the general solution: we fix two successive points $a$, $b=f(a)$ (for example $a=\sqrt{3}/2$, $b=1/2$), choose any (continuous, let's say) increasing $H$ on $[a,b]$ with $H(b)=H(a)+1$, and then the rest is forced on us by (F), by propagating the basic interval by $f$. Conversely, any $H$ obtained in this way will satisfy (F) and $g=1/H'$ will generate a flow with $\phi_1=f$.

Of course, any such $\phi_t$ will automatically be continous (in fact, differentiable) in $t$. However, the remaining two conditions seem tedious (though not impossible perhaps) to analyze in this way.

Your second condition is now equivalent to $H$ being odd about $1/2$, and one can write down more explicit versions, but I'm too lazy now to continue further. It appears that with my above choice of $a$, $b$, one only needs to verify this condition on $[\sqrt{3}/2, 1-\sqrt{3}/2]$, once it holds there, (F) will propagate it to the rest of the interval.

Not really an answer, but maybe slightly more than a comment:

You want to embed the discrete dynamical system $f$ in a flow $\phi_t$ (so $\phi_1=f$), and the most straightforward way to attempt this is to just write down the ODE $\dot{x}=g(x)$ that produces $\phi_t$ and see what conditions $g$ must satisfy. If $H'=1/g$, then the solution is (implicitly) given by $H(\phi_t(x))-H(x)=t$. So in our situation $H$ must satisfy the functional equation $$ H\circ f - H = 1 \quad\quad\quad\quad (F) $$ In principle at least, we can easily describe the general solution: we fix two successive points $a$, $b=f(a)$ (for example $a=\sqrt{3}/2$, $b=1/2$), choose any (continuously differentiable, let's say) increasing $H$ on $[a,b]$ with $H(b)=H(a)+1$, and then the rest is forced on us by (F), by propagating the basic interval by $f$. Conversely, any $H$ obtained in this way will satisfy (F) and $g=1/H'$ will generate a flow with $\phi_1=f$.

Of course, any such $\phi_t$ will automatically be continous (in fact, differentiable) in $t$. However, the remaining two conditions seem tedious (though not impossible perhaps) to analyze in this way.

Your second condition is now equivalent to $H$ being odd about $1/2$, and one can write down more explicit versions, but I'm too lazy now to continue further. It appears that with my above choice of $a$, $b$, one only needs to verify this condition on $[\sqrt{3}/2, 1-\sqrt{3}/2]$, once it holds there, (F) will propagate it to the rest of the interval.

Not really an answer, but maybe slightly more than a comment:

You want to embed the discrete dynamical system $f$ in a flow $\phi_t$ (so $\phi_1=f$), and the most straightforward way to attempt this is to just write down the ODE $\dot{x}=g(x)$ that produces $\phi_t$ and see what conditions $g$ must satisfy. If $H'=1/g$, then the solution is (implicitly) given by $H(\phi_t(x))-H(x)=t$. So in our situation $H$ must satisfy the functional equation $$ H\circ f - H = 1 \quad\quad\quad\quad (F) $$ In principle at least, we can easily describe the general solution: we fix two successive points $a$, $b=f(a)$ (for example $a=\sqrt{3}/2$, $b=1/2$), choose any (continuous, let's say) increasing $H$ with $H(b)=H(a)+1$, and then the rest is forced on us by (F), by propagating the basic interval by $f$. Conversely, any $H$ obtained in this way will satisfy (F) and $g=1/H'$ will generate a flow with $\phi_1=f$.

Of course, any such $\phi_t$ will automatically be continous (in fact, differentiable) in $t$. However, the remaining two conditions seem tedious (though not impossible perhaps) to analyze in this way.

Your second condition is now equivalent to $H$ being odd about $1/2$, and one can write down more explicit versions, but I'm too lazy now to continue further. It appears that with my above choice of $a$, $b$, one only needs to verify this condition on $[\sqrt{3}/2, 1-\sqrt{3}/2]$, once it holds there, (F) will propagate it to the rest of the interval.

Not really an answer, but maybe slightly more than a comment:

You want to embed the discrete dynamical system $f$ in a flow $\phi_t$ (so $\phi_1=f$), and the most straightforward way to attempt this is to just write down the ODE $\dot{x}=g(x)$ that produces $\phi_t$ and see what conditions $g$ must satisfy. If $H'=1/g$, then the solution is (implicitly) given by $H(\phi_t(x))-H(x)=t$. So in our situation $H$ must satisfy the functional equation $$ H\circ f - H = 1 \quad\quad\quad\quad (F) $$ In principle at least, we can easily describe the general solution: we fix two successive points $a$, $b=f(a)$ (for example $a=\sqrt{3}/2$, $b=1/2$), choose any (continuous, let's say) increasing $H$ on $[a,b]$ with $H(b)=H(a)+1$, and then the rest is forced on us by (F), by propagating the basic interval by $f$. Conversely, any $H$ obtained in this way will satisfy (F) and $g=1/H'$ will generate a flow with $\phi_1=f$.

Of course, any such $\phi_t$ will automatically be continous (in fact, differentiable) in $t$. However, the remaining two conditions seem tedious (though not impossible perhaps) to analyze in this way.

Your second condition is now equivalent to $H$ being odd about $1/2$, and one can write down more explicit versions, but I'm too lazy now to continue further. It appears that with my above choice of $a$, $b$, one only needs to verify this condition on $[\sqrt{3}/2, 1-\sqrt{3}/2]$, once it holds there, (F) will propagate it to the rest of the interval.

Not really an answer, but maybe slightly more than a comment:

You want to embed the discrete dynamical system $f$ in a flow $\phi_t$ (so $\phi_1=f$), and the most straightforward way to attempt this is to just write down the ODE $\dot{x}=g(x)$ that produces $\phi_t$ and see what conditions $g$ must satisfy. If $H'=1/g$, then the solution is (implicitly) given by $H(\phi_t(x))-H(x)=t$. So in our situation $H$ must satisfy the functional equation $$ H\circ f - H = 1 \quad\quad\quad\quad (F) $$ In principle at least, we can easily describe the general solution: we fix two successive points $a$, $b=f(a)$ (for example $a=\sqrt{3}/2$, $b=1/2$), choose any (continuous, let's say) increasing $H$ with $H(b)=H(a)+1$, and then the rest is forced on us by (F), by propagating the basic interval by $f$. Conversely, any $H$ obtained in this way will satisfy (F).

Of course, any such $\phi_t$ will automatically be continous (in fact, differentiable) in $t$. However, the remaining two conditions seem tedious (though not impossible perhaps) to analyze in this way.

Your second condition is now equivalent to $H$ being odd about $1/2$, and one can write down more explicit versions, but I'm too lazy now to continue further. It appears that with my above choice of $a$, $b$, one only needs to verify this condition on $[\sqrt{3}/2, 1-\sqrt{3}/2]$, once it holds there, (F) will propagate it to the rest of the interval.

Not really an answer, but maybe slightly more than a comment:

You want to embed the discrete dynamical system $f$ in a flow $\phi_t$ (so $\phi_1=f$), and the most straightforward way to attempt this is to just write down the ODE $\dot{x}=g(x)$ that produces $\phi_t$ and see what conditions $g$ must satisfy. If $H'=1/g$, then the solution is (implicitly) given by $H(\phi_t(x))-H(x)=t$. So in our situation $H$ must satisfy the functional equation $$ H\circ f - H = 1 \quad\quad\quad\quad (F) $$ In principle at least, we can easily describe the general solution: we fix two successive points $a$, $b=f(a)$ (for example $a=\sqrt{3}/2$, $b=1/2$), choose any (continuous, let's say) increasing $H$ with $H(b)=H(a)+1$, and then the rest is forced on us by (F), by propagating the basic interval by $f$. Conversely, any $H$ obtained in this way will satisfy (F) and $g=1/H'$ will generate a flow with $\phi_1=f$.

Of course, any such $\phi_t$ will automatically be continous (in fact, differentiable) in $t$. However, the remaining two conditions seem tedious (though not impossible perhaps) to analyze in this way.

Your second condition is now equivalent to $H$ being odd about $1/2$, and one can write down more explicit versions, but I'm too lazy now to continue further. It appears that with my above choice of $a$, $b$, one only needs to verify this condition on $[\sqrt{3}/2, 1-\sqrt{3}/2]$, once it holds there, (F) will propagate it to the rest of the interval.