A Penrose diagram is a convenient way of representing the causal structure of a given spacetime. Of course, different spacetimes may have equivalent causal structures, as long as their metrics are conformally equivalent. When we work with the Penrose diagram for Minkowski spacetime, we are typically juggling two different spacetimes: Minkowski spacetime $M$ on the one hand, and the "Einstein static universe" $E$ on the other hand. Both have the same causal structure, but their metrics are different: conformally equivalent, but not isometric. Spacetime $M$ is an open subset of spacetime $E$. Let $\overline{M}$ denote the closed subset obtained from $M$ by adjoining the boundary (including $\mathscr{I}^\pm$), so
$$
M\subset\overline{M}\subset E.
$$
The central messages of this answer are:
The Minkowski metric $g$ on $M$ cannot be extended to $\overline{M}$. This follows from the fact that null geodesics which are affinely parameterized with respect to the Minkowski metric on $M$ never reach the boundary for any finite value of the affine parameter, and a coordinate transform cannot change this. (The quality of being affinely parameterized is coordinate-independent.)
On $E$ we have a different metric $\hat g$ that is well-defined everywhere on the closed subset $\overline{M}$, including on the boundary. The metric $\hat g$ is conformally equivalent to $g$ within the subset $M$, so null geodesics in $M$ are also (segments of) null geodesics in $E$. Furthermore, there are (finite segments of) null geodesics within the boundary of $\overline{M}$, as long as "geodesic" is defined with respect to $\hat g$.
Even though $g$ and $\hat g$ are in the same conformal equivalence class and therefore have the same null geodesics wherever both metrics are defined (namely within $M$), they don't give the same affine parameterizations of those null geodesics. This is why we can have a well-defined metric $\hat g$ on $\overline{M}$ which is in the same conformal class as the Minkowski metric $g$ on $M$, even though $g$ itself cannot itself be extended to $\overline{M}$.
When we talk about a geodesic "reaching" a point (on $\mathscr{I}^+$, for example), we are implicitly talking about reaching that point at a finite value of the affine parameter. For example, a spacetime is called "geodesically complete" if every affinely parameterized geodesic can be continued to arbitrarily large values of the affine parameter. The "affine parameter" qualifier is essential, so that Minkowski spacetime can be regarded as geodesically complete.
...it seems a little delicate to set things up mathematically so that null geodesics can get to null infinity but can't have any motion within null infinity. Is there some way that it's encoded into the formalism that we don't have such things going on?
Yes. We simply need to keep track of which metric ($g$ or $\hat g$) we are using to define the affine parameterization. If we are using the Minkowski metric $g$, then null geodesics in $M$ never reach $\mathscr{I}^\pm$; and if we are using the conformally-equivalent metric $\hat g$, then null geodesics in $M$ pass through $\mathscr{I}^\pm$ (which also contains [finite segments of] null geodesics), so $\mathscr{I}^\pm$ isn't really a "boundary" at all unless we arbitrarily choose to ignore everything in $E$ outside of $\overline{M}$.
The rest of this answer makes this mathematically explicit. Instead of starting with Minkowski spacetime $(M,g)$ and constructing its conformal compactification (which is what the Penrose diagram represents), I'll start with the Einstein static universe $(E,\hat g)$, because it's bigger, and then I'll construct $M$ as a topologically open subset of $E$ and review how the metrics $g$ and $\hat g$ are related to each other.
To define the Einstein static universe $(E,\hat g)$, start with the smooth manifold
$$
E=\mathbb{R}\times S^{D-1}
\hskip2cm
\text{(topologically)},
\tag{1}
$$
where $D$ is the number of dimensions. Use a coordinate $T$ to cover the $\mathbb{R}$ factor. For the factor $S^{D-1}$, use the standard metric on the unit sphere, written as
$$
d\Omega_{D-1}^2=dR^2+\sin^2(R)d\Omega_{D-2}^2
\\
0\leq R\leq \pi.
\tag{2}
$$
Intuitively, the coordinate $R$ is analogous to latitude measured from one of the poles, so that the equator is at $R=\pi/2$, and the second term on the right-hand side of (2) is the metric of a constant-latitude slice of $S^{D-1}$, which has topology $S^{D-2}$ and radius $\sin(R)$. The Einstein static universe $E$ is defined to be the manifold (1) equipped with the metric $\hat g$ defined by
$$
\hat g=dT^2-\left(dR^2+\sin^2(R)d\Omega_{D-2}^2\right)
\\
-\infty<T<\infty
\hskip2cm
0\leq R\leq \pi.
\tag{3}
$$
The spacetime $(E,\hat g)$ is geodesically complete, geometrically curved, and topologically non-trivial (not homeomorphic to $\mathbb{R}^D$). This is the Einstein static universe.
Now we can construct $D$-dimensional Minkowski spacetime $(M,g)$ by taking a particular open subset $M\subset E$ and giving it a new metric $g$. Within $M$, the new metric $g$ will be conformally equivalent to $\hat g$, so null geodesics in $M$ still coincide with (segments of) null geodesics in $E$. This important fact is stated with exceptional clarity in [$1$]:
Two metrics related by an overall scalar multiple function are said to be "conformally related," or related by a "Weyl rescaling" or "Weyl transformation." The light cones of two such metrics $g_{\mu\nu}$ and $A^2(x)g_{\mu\nu}$ are obviously the same, and hence so are the null curves. ... in fact the null geodesic curves are also the same, but ... the affine parameters on those curves are not the same.
(I italicized the part that is central to this answer.) To construct $M$, consider the topologically-open subset of $E$ defined by
$$
0\leq |T|+R<\pi
\tag{4}.
$$
This subset is nicely illustrated in the following figure from [$2$]:
In this figure, $\mathbb{R}\times S^{D-1}$ is depicted as $\mathbb{R}\times S^1$, so each horizontal circle really represents a sphere $S^{D-1}$. The subset of $E$ defined by (4) corresponds to the shaded region in the figure. This subset will be denoted $M$. It is topologically trivial (homeomorphic to $\mathbb{R}^D$), because a neighborhood of the point $R=\pi$ is omitted from each of the spheres $S^{D-1}$ that it intersects. (At $T=0$, only the point $R=\pi$ itself is omitted.) The subset $M$ by itself is represented by the following Penrose diagram, which is from [$3$]:
This is half of the diamond-shaped region from the preceding figure, the two halves differing only in the values of the angular coordinates $\Omega_{D-2}$. Note that the symbols $\mathcal{I}^\pm$ in these pictures mean the same thing as the symbols $\mathscr{I}^\pm$ in this answer (and in the OP).
We could continue using the curved metric (3) on $M$, but then $M$ would not be geodesically complete: some null geodesics in $(E,\hat g)$ pass through $(M,\hat g)$ but are not completely contained in $M$. We can make $M$ geodesically complete by replacing the metric (3) with the new metric $g$ defined by
$$
g=\frac{dT^2-\left(dR^2+\sin^2(R)d\Omega_{D-2}^2\right)}{
\cos^2 U\cos^2 V}
\hskip1cm
0\leq |T|+R<\pi
\tag{5}
$$
with $U,V$ being abbreviations for $(T\pm R)/2$. This metric is well-defined everywhere on $M$, where the denominator is non-zero, but it is not well-defined everywhere on $E$. In particular, the denominator goes to zero on the boundary of $\overline{M}$. In particular, it goes to zero on $\mathscr{I}^\pm$. That's okay, because the manifold $M$ does not include $\mathscr{I}^\pm$ due to the restriction $|T|+R<\pi$.
Here is the central point: With respect to the metric $\hat g$ defined by (3), we have null geodesics inside $\mathscr{I}^\pm$, namely any worldline for which the angular coordinates $\Omega_{D-2}$ are constant and $|T|+R=\pi$. However, we cannot say whether or not these worldlines are "geodesics" with respect to the metric $g$ defined by (5) (or (9), below), because $g$ is not defined on $\mathscr{I}^\pm$.
Despite appearances, the metric $g$ defined by (5) is isometric to (not just conformally equivalent to) to the usual Minkowski metric. In particular, the metric (5) describes flat spacetime. To make this explicit, define new coordinates $t,r$ by
$$
t=u+v
\hskip2cm
r=u-v
\tag{6}
$$
with
$$
u=\tan U
\hskip2cm
v=\tan V.
\tag{7}
$$
These equations are wel-defined for all $0\leq U,V < \pi/2$, so the coordinates $t,r$ cover all of $M$, along with the angular coordinates that are collectively denoted $\Omega_{D-2}$ in (5). The corresponding ranges are
$$
-\infty<t<\infty
\hskip2cm
0\leq r<\infty.
\tag{8}
$$
By taking the differential of (6), we can relate $dt,dr$ to $dT,dR$. Using those relationships in (5) gives the Minkowski metric
$$
g=dt^2-(dr^2+r^2 d\Omega_{D-2}^2).
\tag{9}
$$
Within the subset $M$, the new metric (9) is conformally equivalent to the original metric (3), even though it is geometrically different. According to the excerpt that was highlighted above, this means that each null geodesic in $(M,g)$ is a (segment of a) null geodesic in $(E,\hat g)$. However, given a null geodesic, the quality of being affinely parameterized depends on which metric we use. A null geodesic that is affinely parameterized according to $\hat g$ is not necessarily affinely parameterized according to $g$, and conversely. A null geodesic that is affinely parameterized with respect to the Minkowski metric (9) never reaches the boundary $\mathscr{I}^+$ for any finite value of the affine parameter, whereas the same null geodesic affinely parameterized with respect to the metric (3) does reach the boundary at a finite value of the affine parameter, and it continues right through the boundary to the other side, which doesn't even exist in $M$. Furthermore, the boundary itself contains (finite segments of) null geodesics in $E$, whereas $M$ does not include the boundary at all.
In summary, the situation described in the OP never occurs. It does not occur in $(M,g)$ because $(M,g)$ is a geodesically-complete spacetime that excludes $\mathscr{I}^+$, and affinely-parameterized null geodesics in $(M,g)$ don't reach $\mathscr{I}^+$. It does not occur in $(\overline{M},\hat g)$ either, because null geodesics in $(\overline{M},\hat g)$ pass right through $\mathscr{I}^+$ to the other side in $(E,\hat g)$.
References:
[$1$] Jacobson (2007), "HW#3—Phys 675—Fall 2007," http://www.physics.umd.edu/grt/taj/675d/675dhw3.pdf
[$2$] Felinska (2010?), "Carter-Penrose diagrams and black holes
," http://www.ift.uni.wroc.pl/~blaschke/master/Felinska.pdf
[$3$] Aretakis (2013), "Lecture Notes on General Relativity," https://web.math.princeton.edu/~aretakis/columbiaGR.pdf
