[Collecting my sporadic comments into 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 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 :)).
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.
[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.
