Five-fold symmetric Sierpinski gasket pentagon using Iterated Functional System Ran across an old article from Computer Graphics World by Michael F. Barnsley and Alan D. Sloan entitled Chaotic Compression on Iterated Functional Systems and modified their Table 1 $W$ matrix by adding two lines with attractor points separated by $2\pi /5$ rather than $2\pi /3$ and arrived at this Sierpinski pentagon (20000+ iterations of new data point being halfway in between previous data point and a randomly selected (equally weighted) pentagon vertex).

The question is: for this fractal, are there intact miniature copies of itself at all scales?  How does this tie in with incompatibility of five-fold symmetry and plane-tiling?
update
if instead of halving the distance to a randomly selected vertex, I third it or quarter it, the constituent nested pentagons become separated/distinct

 A: The Chaos Game
The image in the question appears to have been rendered using the chaos game technique.  This is a technique for rendering the attractors of iterated function systems.  In a nutshell, start with a collection of self-similar contractions, i.e. maps of the form
$$ \mathbb{R}^d \to \mathbb{R}^d : x \mapsto rx + b, $$
where $r \in (0,1)$ is the contraction ratio and $b\in \mathbb{R}^d$ is a translation.  Given a collection $\{\varphi_j\}_{j=1}^{J}$ of such maps (called an iterated function system or IFS), there is a unique, nonempty, compact set $A$ with the property that
$$ A = \bigcup_{j=1}^{J} \varphi_j(A). $$
This set is called the attractor of the IFS.  Note that each map $\varphi_j$ has a fixed point; call this point $p_j$.  To render an image of $A$ using the chaos game, pick an initial point $P_0$.  Then, given a point $P_n$, choose a successor point $P_{n+1}$ by choosing one of the $p_j$ at random, then setting $P_{n+1}$ to be the $r$-weighted average of $P_n$ and the randomly chosen fixed point.  That is, define
$$ P_{n+1} = r P_n + (1-r)\operatorname{rand}\{ p_1, \dotsc, p_J \}), $$
where $\operatorname{rand} S$ means "choose an element from $S$ randomly with uniform probability".  Observe that if $r = \frac{1}{2}$, then $P_{n+1}$ is just the midpoint of $P_n$ and the randomly chosen point.
For any $\varepsilon > 0$, there exists some $N$ such that
$$ d(P_n, A) = \min_{x\in A} d(P_n,x) < \varepsilon $$
whenever $n > N$ (note that $A$ is compact, so this minimum exists).  This means that the iteratively defined points $P_n$ are eventually very close to the attractor $A$, so we render those points.
The $\frac{1}{2}$-Pentagasket
The image in the question appears to be a rendering of the attractor of the IFS $\{ \varphi_j \}_{j=0}^{4}$, where the maps are given by something like
$$ \varphi_j(x) = \frac{1}{2} x + \frac{1}{2}\exp(ij2\pi/5), $$
where $\exp$ is the complex exponential and $i$ is the imaginary unit, defined by the property that $i^2 = -1$.  In this context, the real plane $\mathbb{R}^2$ is identified with the complex plane $\mathbb{C}$.  Note that the fixed point of $\varphi_j$ is $\exp(ij2\pi/5)$, i.e. a 5th root of unity.  My own rendering of this fractal looks like this (with 10,000,000 iterations, which is probably overkill, but can be done in less than two seconds, so why not):

Note that my version is rotated, as I have taken the fixed points to be the actual 5th roots of unity, while the original image rotates the fixed points by (for example) a quarter of a turn.  However, all of the important properties are the same.
For the sake of naming this object, I am going to call it the $\frac{1}{2}$-pentagasket.  The $\frac{1}{2}$ here is a parameter which I am going to change in am moment, while the phrase "pentagasket" is meant to invoke the idea of the Sierpinski gasket (or Sierpinski triangle), but with more fixed points.
At this point, we can answer the question "for this fractal, are there intact miniature copies of itself at all scales?"  The answer is "Yes!" (depending on what is meant by "intact".  This can be more easily seen in the following image:

The $\frac{1}{2}$-pentagasket is made up of five copies of itself, each of has been scaled by a factor of $\frac{1}{2}$.  However, these mini-gaskets do not "tile" the pentagasket.  Instead, they overlap significantly.
A Spectrum of Pentagaskets
This does not specifically address the question, but I infer that the asker would might be interested in a pentagasket which does tile itself, i.e. is made up entirely of smaller copies of itself which do not overlap.  It turns out that this is related to the contraction ratio.  If the contraction ratio is smaller, then the self-similar copies of the gasket are smaller, and so they needn't overlap.  The critical number turns out to be a contraction ratio of
$$ r = \frac{1}{2\cos(72^{\circ}) + 2}. $$
A good exercise for the reader would be to determine why this is the magic number.  If we retain the same fixed points as above and render an $r$-pentagasket with this value of $r$, we get

This pentagasket is composed of exactly 5 smaller copies of itself which do not overlap, but which do "touch" at the corners.  It is worth noting that this ratio represents the cutoff at which the IFS satisfies the open set condition.  For this and smaller values of $r$, the attractor will have Hausdorff dimension equal to
$$ \dim_{\text{H}} A = - \frac{\log(5)}{\log(r)}. $$
For larger values of $r$, there will be some overlap—the Hausdorff dimension is no longer guaranteed to be equal to the similarity dimension, and more delicate computations may be necessary.  By the time $r=\frac{1}{2}$, the attractor "fills up space" in certain regions, hence one would expect the Hausdoff dimension of the $\frac{1}{2}$-pentagasket to be $2$.
Also, just for giggles, an animation of $r$-pentagaskets with $r$ going from $0.10$ to $0.60$, with a step size of $0.02$:

