This condition is quite useful, because many forgetful functors are monadic, and monadic functors create all limits (by combining their definition on pp. 143--144 of Mac Lane and Ex. 6.2.2 on p. 142 of Mac Lane, or by Proposition 4.4.1 on p. 178 of Mac Lane--Moerdijk, or really by a commentcomment of Tom Leinster from which I learned this :)).
This condition is quite useful, because many forgetful functors are monadic, and monadic functors create all limits (by combining their definition on pp. 143--144 of Mac Lane and Ex. 6.2.2 on p. 142 of Mac Lane, or by Proposition 4.4.1 on p. 178 of Mac Lane--Moerdijk, or really by a comment of Tom Leinster from which I learned this :)).
This condition is quite useful, because many forgetful functors are monadic, and monadic functors create all limits (by combining their definition on pp. 143--144 of Mac Lane and Ex. 6.2.2 on p. 142 of Mac Lane, or by Proposition 4.4.1 on p. 178 of Mac Lane--Moerdijk, or really by a comment of Tom Leinster from which I learned this :)).
[Collecting my sporadic comments tointo one (hopefully) coherent answer.]
This condition is quite useful, because many forgetful functors are monadic, and monadic functors create all limits (by combining their definition on pp. 143--144 of Mac Lane and by Ex. 6.2.2 on p. 142 of Mac Lane, or by Proposition 4.4.1 on p. 178 of Mac Lane--Moerdijk, or really by a comment of Tom Leinster from which I learned this :)).
Regarding your example in Q. 23188: Unfortunately I know nothing of Hopf algebras, so I can't understand all the details of your construction. If I understand correctly, you construct a functor $F\colon\mathbf{Rng}\to\mathbf{Grp}$ whose composition with the forgetful functor $U\colon \mathbf{Grp}\to \mathbf{Set}$ is representable. If this is indeed the case, then by the above $F$ itself preserves all limits.
[EDIT: corrected the part concerning the last example.]
Finally, regarding your example in the edited question: While While I know nothing of dynamical systems, from a quick glance at Terence Tao's blog it seems to me that these are just algebras for the trivial monadcategory of dynamical systems is the $\mathbb{T}=\langle \mathrm{Id}_{\mathbf{Set}},1,1\rangle$category whose objects are pairs $\langle X,f\colon X\to X\rangle$ with $X$ a (small) set and whose arrows $\phi\colon\langle X, f\rangle\to\langle Y, g\rangle$ are those functions $\phi\colon X\to Y$ with $g\circ\phi =\phi\circ f$.
To show that the above sufficient condition works in this case, we would like to show that the forgetful functor to $\mathbf{Set}$ crates limits. So More generally, we will show that if in your example you construct$C$ is a functor $F\colon\mathbf{Set}\to\mathbf{Set}^{\mathbb{T}}$ whose composition withcategory and $D$ is the category whose objects are pairs $\langle x,f\colon x\to x\rangle$ (where $x\in\operatorname{obj}(C)$, forgetful$f\in\operatorname{arr}(C)$), and whose arrows $\phi\colon \langle x,f\rangle\to \langle y,g\rangle$ are those arrows $\phi\colon x\to y$ with $g\circ\phi =\phi\circ f$, then the forgetful functor $G^{\mathbb{T}}\colon \mathbf{Set}^{\mathbb{T}}\to\mathbf{Set}$ preserves$U\colon D\to C$ creates limits.
[I'm sure that this follows from some well-known result, then yourbut since I $F$ preserves limitsdon't see it, I'll just continue with a direct proof.]
So, let $J$ be an index category, let
$F\colon J\to D$ be a functor, and suppose that $\tau\colon
x\stackrel{.}{\to} UF$ is a limiting cone in $C$. We would like to
show that there exists a
unique cone $\sigma\colon L\stackrel{.}{\to} F$ in $D$ such that
$U\sigma=\tau$, and that this unique cone is a limiting cone.
For uniqueness, suppose that $\sigma\colon L\stackrel{.}{\to} F$ satisfies $U\sigma = \tau$. Write $F_j:=\langle y_j,f_j\rangle$. Then we must have for all $j$ $$ \sigma_j=(x\stackrel{f}{\to}x)\stackrel{\tau_j}{\to}(y_j\stackrel{f_j}{\to}y_j) $$ for some $f\colon x\to x$ (by Exhence we immediately see that $\sigma$ is determined up to $f$). 6Now, since by the above we see that $\tau_j$ must be an arrow $$ (x\stackrel{f}{\to}x)\stackrel{\tau_j}{\to}(y_j\stackrel{f_j}{\to}y_j) $$ of $D$, the following diagram must be commutative for all $j$: $$ \begin{matrix} x & \stackrel{\tau_j}{\longrightarrow} & y_j =UF_j\\ f\downarrow & & f_j\downarrow\\ x&\stackrel{\tau_j}{\longrightarrow} & y_j = UF_j. \end{matrix} \quad \text{(Diagram 1)} $$
Now we claim that the $\to\downarrow$ part of the above diagram forms a cone to $UF$, that is, we claim that the family $\{f_j\tau_j\}$ forms a cone $x\stackrel{.}{\to} UF$.2 Indeed, for an arrow $g:j\to j'$ of $J$, consider the following diagram: $$ \begin{matrix} &&&&x\\ &&&\stackrel{\tau_j}{\swarrow}&&\stackrel{\tau_{j'}}{\searrow}\\ &&y_j && \stackrel{UFg}{\longrightarrow} && y_{j'}\\ &\stackrel{f_j}{\swarrow} &&&&&&\stackrel{f_{j'}}{\searrow}\\ y_j&&&&\stackrel{UFg}{\longrightarrow}&&&&y_{j'} \end{matrix} $$
The upper triangle is commutative because $\tau$ is a cone to the base $UF$, and the lower trapezoid is commutative because $F$ is a functor, and hence $Fg$ is an arrow $F_j\to F_{j'}$ in $D$.2 Hence the outer triangle commutes, pas required. 142From the universality of ML)$\tau$, it follows that there is a unique $f$ for which Diagram 1 is commutative, and we have uniqueness.
For existence, we can take $f$ to be the unique arrow $x\to x$ for which Diagram 1 is commutative, and we get a cone $$ \sigma=\{\sigma_j=\tau_j\colon (x\stackrel{f}{\to}x)\to F_j=(y_j\stackrel{f_j}{\to}y_j)\} $$ with $U\sigma=\tau$. We claim that this is a limiting cone.
To see this, let $\alpha\colon(z\stackrel{g}{\to}z)\stackrel{.}{\to}F$ be a cone, so that for all $j$ the following diagram is commutative: $$ \begin{matrix} z & \stackrel{\alpha_j}{\longrightarrow} & y_j\\ g\downarrow & & f_j\downarrow\\ z &\stackrel{\alpha_j}{\longrightarrow} & y_j. \end{matrix} \quad\text{(Diagram 2)} $$
Then $U\alpha$ is a cone $z\stackrel{.}{\to} UF$ in $C$, and by the universality of $\tau$ there exists a unique arrow $h\colon z\to x$ for which the following diagram is commutative for all $j$: $$ \begin{matrix} z & \stackrel{\alpha_j}{\longrightarrow} & y_j\\ h\downarrow& \stackrel{\tau_j}{\nearrow}\\ x& \end{matrix}\quad\text{(Diagram 3)} $$
If this $h$ is an arrow $(z\stackrel{g}{\to}z)\to (x\stackrel{f}{\to}x)$ in $D$, then we're done. In other words, all that remains to do is to show that the outer rectangle of the following diagram is commutative: $$ \begin{matrix} z && \stackrel{h}{\longrightarrow} && x\\ & \stackrel{\alpha_j}{\searrow} && \stackrel{\tau_j}{\swarrow}\\ && y_j\\ g\downarrow&& \downarrow f_j && \downarrow f\\ && y_j\\ & \stackrel{\alpha_j}{\nearrow} && \stackrel{\tau_j}{\nwarrow}\\ z && \stackrel{h}{\longrightarrow} && x\\ \end{matrix} $$ Now, the left trapezoid is just Diagram 2, the upper and lower triangles are just Diagram 3, and the right trapezoid is commutative for all $j$ by the definition of $f$. It follows that both paths of the outer rectangle have the same composition with the limiting cone $\tau$, and hence the outer rectangle is commutative, as required.
[Collecting my sporadic comments to one (hopefully) coherent answer.]
This condition is quite useful, because many forgetful functors are monadic, and monadic functors create all limits (by their definition on pp. 143--144 of Mac Lane and by Ex. 6.2.2 on p. 142 of Mac Lane, or by Proposition 4.4.1 on p. 178 of Mac Lane--Moerdijk, or really by a comment of Tom Leinster from which I learned this :)).
Regarding your example in Q. 23188: Unfortunately I know nothing of Hopf algebras, so I can't understand all the details of your construction. If I understand correctly, you construct a functor $F\colon\mathbf{Rng}\to\mathbf{Grp}$ whose composition with the forgetful functor $U\colon \mathbf{Grp}\to \mathbf{Set}$ is representable. If this is indeed the case, then by the above $F$ itself preserves all limits.
Finally, regarding your example in the edited question: While I know nothing of dynamical systems, from a quick glance at Terence Tao's blog it seems to me that these are just algebras for the trivial monad $\mathbb{T}=\langle \mathrm{Id}_{\mathbf{Set}},1,1\rangle$ in $\mathbf{Set}$. So, if in your example you construct a functor $F\colon\mathbf{Set}\to\mathbf{Set}^{\mathbb{T}}$ whose composition with the forgetful functor $G^{\mathbb{T}}\colon \mathbf{Set}^{\mathbb{T}}\to\mathbf{Set}$ preserves limits, then your $F$ preserves limits (by Ex. 6.2.2, p. 142 of ML).
[Collecting my sporadic comments into one (hopefully) coherent answer.]
This condition is quite useful, because many forgetful functors are monadic, and monadic functors create all limits (by combining their definition on pp. 143--144 of Mac Lane and Ex. 6.2.2 on p. 142 of Mac Lane, or by Proposition 4.4.1 on p. 178 of Mac Lane--Moerdijk, or really by a comment of Tom Leinster from which I learned this :)).
Regarding your example in Q. 23188: Unfortunately I know nothing of Hopf algebras, so I can't understand all the details of your construction. If I understand correctly, you construct a functor $F\colon\mathbf{Rng}\to\mathbf{Grp}$ whose composition with the forgetful functor $U\colon \mathbf{Grp}\to \mathbf{Set}$ is representable. If this is indeed the case, then by the above $F$ itself preserves all limits.
[EDIT: corrected the part concerning the last example.]
Finally, regarding your example in the edited question: While I know nothing of dynamical systems, from a quick glance at Terence Tao's blog it seems that the category of dynamical systems is the category whose objects are pairs $\langle X,f\colon X\to X\rangle$ with $X$ a (small) set and whose arrows $\phi\colon\langle X, f\rangle\to\langle Y, g\rangle$ are those functions $\phi\colon X\to Y$ with $g\circ\phi =\phi\circ f$.
To show that the above sufficient condition works in this case, we would like to show that the forgetful functor to $\mathbf{Set}$ crates limits. More generally, we will show that if $C$ is a category and $D$ is the category whose objects are pairs $\langle x,f\colon x\to x\rangle$ (where $x\in\operatorname{obj}(C)$, $f\in\operatorname{arr}(C)$), and whose arrows $\phi\colon \langle x,f\rangle\to \langle y,g\rangle$ are those arrows $\phi\colon x\to y$ with $g\circ\phi =\phi\circ f$, then the forgetful functor $U\colon D\to C$ creates limits.
[I'm sure that this follows from some well-known result, but since I don't see it, I'll just continue with a direct proof.]
So, let $J$ be an index category, let
$F\colon J\to D$ be a functor, and suppose that $\tau\colon
x\stackrel{.}{\to} UF$ is a limiting cone in $C$. We would like to
show that there exists a
unique cone $\sigma\colon L\stackrel{.}{\to} F$ in $D$ such that
$U\sigma=\tau$, and that this unique cone is a limiting cone.
For uniqueness, suppose that $\sigma\colon L\stackrel{.}{\to} F$ satisfies $U\sigma = \tau$. Write $F_j:=\langle y_j,f_j\rangle$. Then we must have for all $j$ $$ \sigma_j=(x\stackrel{f}{\to}x)\stackrel{\tau_j}{\to}(y_j\stackrel{f_j}{\to}y_j) $$ for some $f\colon x\to x$ (hence we immediately see that $\sigma$ is determined up to $f$). Now, since by the above we see that $\tau_j$ must be an arrow $$ (x\stackrel{f}{\to}x)\stackrel{\tau_j}{\to}(y_j\stackrel{f_j}{\to}y_j) $$ of $D$, the following diagram must be commutative for all $j$: $$ \begin{matrix} x & \stackrel{\tau_j}{\longrightarrow} & y_j =UF_j\\ f\downarrow & & f_j\downarrow\\ x&\stackrel{\tau_j}{\longrightarrow} & y_j = UF_j. \end{matrix} \quad \text{(Diagram 1)} $$
Now we claim that the $\to\downarrow$ part of the above diagram forms a cone to $UF$, that is, we claim that the family $\{f_j\tau_j\}$ forms a cone $x\stackrel{.}{\to} UF$. Indeed, for an arrow $g:j\to j'$ of $J$, consider the following diagram: $$ \begin{matrix} &&&&x\\ &&&\stackrel{\tau_j}{\swarrow}&&\stackrel{\tau_{j'}}{\searrow}\\ &&y_j && \stackrel{UFg}{\longrightarrow} && y_{j'}\\ &\stackrel{f_j}{\swarrow} &&&&&&\stackrel{f_{j'}}{\searrow}\\ y_j&&&&\stackrel{UFg}{\longrightarrow}&&&&y_{j'} \end{matrix} $$
The upper triangle is commutative because $\tau$ is a cone to the base $UF$, and the lower trapezoid is commutative because $F$ is a functor, and hence $Fg$ is an arrow $F_j\to F_{j'}$ in $D$. Hence the outer triangle commutes, as required. From the universality of $\tau$, it follows that there is a unique $f$ for which Diagram 1 is commutative, and we have uniqueness.
For existence, we can take $f$ to be the unique arrow $x\to x$ for which Diagram 1 is commutative, and we get a cone $$ \sigma=\{\sigma_j=\tau_j\colon (x\stackrel{f}{\to}x)\to F_j=(y_j\stackrel{f_j}{\to}y_j)\} $$ with $U\sigma=\tau$. We claim that this is a limiting cone.
To see this, let $\alpha\colon(z\stackrel{g}{\to}z)\stackrel{.}{\to}F$ be a cone, so that for all $j$ the following diagram is commutative: $$ \begin{matrix} z & \stackrel{\alpha_j}{\longrightarrow} & y_j\\ g\downarrow & & f_j\downarrow\\ z &\stackrel{\alpha_j}{\longrightarrow} & y_j. \end{matrix} \quad\text{(Diagram 2)} $$
Then $U\alpha$ is a cone $z\stackrel{.}{\to} UF$ in $C$, and by the universality of $\tau$ there exists a unique arrow $h\colon z\to x$ for which the following diagram is commutative for all $j$: $$ \begin{matrix} z & \stackrel{\alpha_j}{\longrightarrow} & y_j\\ h\downarrow& \stackrel{\tau_j}{\nearrow}\\ x& \end{matrix}\quad\text{(Diagram 3)} $$
If this $h$ is an arrow $(z\stackrel{g}{\to}z)\to (x\stackrel{f}{\to}x)$ in $D$, then we're done. In other words, all that remains to do is to show that the outer rectangle of the following diagram is commutative: $$ \begin{matrix} z && \stackrel{h}{\longrightarrow} && x\\ & \stackrel{\alpha_j}{\searrow} && \stackrel{\tau_j}{\swarrow}\\ && y_j\\ g\downarrow&& \downarrow f_j && \downarrow f\\ && y_j\\ & \stackrel{\alpha_j}{\nearrow} && \stackrel{\tau_j}{\nwarrow}\\ z && \stackrel{h}{\longrightarrow} && x\\ \end{matrix} $$ Now, the left trapezoid is just Diagram 2, the upper and lower triangles are just Diagram 3, and the right trapezoid is commutative for all $j$ by the definition of $f$. It follows that both paths of the outer rectangle have the same composition with the limiting cone $\tau$, and hence the outer rectangle is commutative, as required.
[Collecting my sporadic comments to one (hopefully) coherent answer.]
A more general question is as follows: For functors $C\stackrel{F}{\to}D\stackrel{U}{\to}E$ and for an index category $J$ such that $UF$ preserves $J$-limits, when does $F$ preserve $J$ limits?
A useful sufficient condition is that if $U$ creates $J$-limits, then in the above situation $F$ preserves $J$-limits. Proof: Let $T\colon J\to C$ be a functor, and suppose that $\tau\colon \ell\stackrel{\cdot}{\to} T$ is a limiting cone in $C$. Since $UF$ preserves $J$-limits, $UF\tau\colon UF\ell\stackrel{\cdot}{\to} UFT$ is a limiting cone in $E$. As $U$ creates $J$-limits, there is a unique lifting of $UF\tau$ to a cone in $D$, and this cone is a limiting cone. But $F\tau\colon F\ell\stackrel{\cdot}{\to} FT$ is such a lift, and hence we're done.
This condition is quite useful, because many forgetful functors are monadic, and monadic functors create all limits (by their definition on pp. 143--144 of Mac Lane and by Ex. 6.2.2 on p. 142 of Mac Lane, or by Proposition 4.4.1 on p. 178 of Mac Lane--Moerdijk, or really by a comment of Tom Leinster from which I learned this :)).
For example, consider the category of all small algebraic systems of some type. From the AFT, we know that the forgetful functor to $\mathbf{Set}$ has a left adjoint, and it is the content of Theorem 6.8.1, p. 156 of Mac Lane that this forgetful functor is monadic.
Returning to the original question, this means that whenever the category $D$ is one of $\mathbf{Grp}$, $\mathbf{Rng}$, $\mathbf{Ab}$,... and $U\colon D\to \mathbf{Set}$ is the forgetful functor, then for any $J$, $UF$ preserves $J$-limits implies $F$ preserves $J$ limits. In particular, if $UF$ is a representable functor (and hence preserves all limits), then $F$ preserves all limits.
Next, let me try to comment on your motivating examples (the one from Q. 23188 and the one from the 'Edit' part of the current question.)
Regarding your example in Q. 23188: Unfortunately I know nothing of Hopf algebras, so I can't understand all the details of your construction. If I understand correctly, you construct a functor $F\colon\mathbf{Rng}\to\mathbf{Grp}$ whose composition with the forgetful functor $U\colon \mathbf{Grp}\to \mathbf{Set}$ is representable. If this is indeed the case, then by the above $F$ itself preserves all limits.
Finally, regarding your example in the edited question: While I know nothing of dynamical systems, from a quick glance at Terence Tao's blog it seems to me that these are just algebras for the trivial monad $\mathbb{T}=\langle \mathrm{Id}_{\mathbf{Set}},1,1\rangle$ in $\mathbf{Set}$. So, if in your example you construct a functor $F\colon\mathbf{Set}\to\mathbf{Set}^{\mathbb{T}}$ whose composition with the forgetful functor $G^{\mathbb{T}}\colon \mathbf{Set}^{\mathbb{T}}\to\mathbf{Set}$ preserves limits, then your $F$ preserves limits (by Ex. 6.2.2, p. 142 of ML).