This is an expansion and follow up of my comment.
As mentioned in comment.
$\hspace0.5in$ Without further restriction, there are no relation.
An example is the function $f(x) = x^3$ which is invertible over the real axis and yet its inverse function doesn't have a power series expansion at $x = 0$.
In general, if your function is invertible only over real axis (or part of it), there isn't much one can do but check the power expansions of the function and its inverse function separately.
However, if your function is analytic and injective over some open subset of $\mathbb{C}$ (such a function is known as an univalent function), there is a theorem by Koebe which can help you.
Koebe quarter theorem
The image of any univalent function $\varphi : B(0,1) \to \mathbb{C}$ from the unit disk $B(0,1)$ onto a subset of the complex plane contains the disk $\displaystyle\;B\left( \varphi(0), \frac{|\varphi'(0)|}{4}\right)\;$.
Let's say your function $f$ is univalent over the open disk $B(z_0,R)$. Since $f$ is analytic and locally injective at $z = z_0$, $f'(z_0) \ne 0$. Now define a function $\varphi : B(0,1) \to \mathbb{C}$ by
$$B(0,1) \ni \omega \mapsto \varphi(\omega) = f(z_0 + R \omega ) \in \mathbb{C}$$
We have $\varphi(0) = f(z_0)$ and $\varphi'(0) = R f'(z_0)$. By Koebe quarter theorem,
$$f(B(z_0,R)) = \varphi(B(0,1)) \supset B\left( f(z_0), \frac{R|f'(z_0)|}{4} \right)$$
This implies the nearest singularity of the inverse function $\varphi^{-1}$ in $\mathbb{C}$ is at least at a distance $\displaystyle\;\frac{R|f'(z_0)|}{4}$ from $f(z_0)$. As a result, the radius of convergence of the power series of $\varphi^{-1}$ and hence that of $f^{-1}$ at $f(z_0)$ is at least $\displaystyle\;\frac{R|f'(z_0)|}{4}$.
I'm having some difficulty parsing your exact intent. (It's early here; perhaps too early.) So I'll write some true things and then we can see what happens...
Let $C$ be a circle; $C = \{z \in \Bbb{C} : |z-c_C| = r_C\}$, where $c_C$ and $r_C$ are the center and radius of $C$, respectively.
On $\Bbb{C}$, a power series converges either on all of $\Bbb{C}$ (any entire function, like $\mathrm{e}^x$), on an open disk of some radius ($1/z$ expanded around any point other than $0$), or on a point ($\sum_{n=0}^\infty n! z^n$). If the region of convergence is $\Bbb{C}$ or an open disk, then on $C$, the region of convergence is some (relatively) open set in $C$. In this case, any point of convergence in $C$ has an open neighborhood of convergence in $C$. If the region of convergence is a point, then either the point is on $C$ or it isn't. In either case the region of convergence is some (relatively) closed set in $C$ and, if that point is in $C$, the only subset of $C$ containing the point of convergence is the point of convergence.
Regarding your last idea: Since $f$ is given by a power series, it is analytic. It is a theorem that if an analytic function has an accumulation point in any of its level sets it is constant. (Consider a sequence of points approaching the accumulation point with all points chosen from the level set. Using these points as a particular sequence in the limit definition of the derivative, we find the derivative at the accumulation point is zero. Repeating for higher derivatives, the same thing happens for them. Consequently, expanding a new power series at the accumulation point, all the derivatives are zero, so the function is a constant, and since this point was in the disk of convergence of the original power series, the two power seriess agree on a disk centered at the accumulation point, so define the same function. I.e., the function was always constant. You have the same problem (only without convergence) if the set of points of divergence is dense at some accumulation point. To see this, consider the function $1/f$ (whose zero level set now has an accumulation point, so this function is zero).
Edit: So the question was potentially about the boundary of the disk of convergence. (Although this was never stated. The circle and the boundary of the disk of convergence were never related in the OP.) On that circle, there is at least one point where the power series cannot be analytically continued (otherwise, the radius of convergence is actually larger). Note that this is not the same as a point of divergence. ($\displaystyle \sum_{n=1}^\infty \frac{z^n}{n^2}$ converges on its entire boundary.)
You might head over to this MathOverflow question. The first answer is thorough and provides a few nice references. The short answer is that the set of convergence on the circle is a countable intersection of countable unions of closed sets. It is conjectured that the sets of possible convergence are exactly the collection of countable intersections of countable unions of closed sets. (So just because we can write down such a set doesn't automatically make it the set of convergence for some power series.)
I'm not convinced your question has a settled answer at present.
Best Answer
You didn't understand what the Wikipedia article says. It says the IF the series $f(X)$ converges, THEN the limit is a polynomial in $X$. It doesn't always converge. For example, take $f(X) = \sum_{k \geqslant 0} X^k$ and $X = \lambda I_n$ a homethety, then $f(X) = \left(\sum_{k \geqslant 0} \lambda^k\right)I_n$ converges if and only if $|\lambda| < 1$.
What says you Korean article is almost right however. Let $(a_n)$ be a sequence of complex numbers and $R$ be its convergence radius. Let $X$ be an $n \times n$ matrix with complex coefficients and $\lambda_1,\ldots,\lambda_n$ be its eigenvalues counted with multiplicities and $\rho = \max\{|\lambda_k|\}$ be its spectral radius. You can prove that $\rho < R \Rightarrow f(X)$ converges $\Rightarrow \rho \leqslant R$ and the case $\rho = R$ is complicated, just like for classical complex series.
You can prove the second statement by contraposition. Let $v_i$ be a non zero vector such that $Xv_i = \lambda_iv_i$ with $|\lambda_i| > R$. Then $f(\lambda_i)$ diverges hence $$ \left(\sum_{k = 0}^K a_kX^k\right)v_i = \left(\sum_{k = 0}^K a_k\lambda_i^k\right)v_i \textrm{ diverges when $K \rightarrow +\infty$}, $$ so $f(X)$ diverges.
For the first statement, begin by the simple case where $X = P^{-1}DP$ is diagonalisable. We have $D = \mathrm{diag}(\lambda_1,\ldots,\lambda_n)$ and, \begin{align*} \sum_{k = 0}^K a_kX^k & = \sum_{k = 0}^K a_kPD^kP^{-1}\\ & = P\left(\sum_{k = 0}^K a_k\mathrm{diag}(\lambda_1^k,\ldots,\lambda_n^k)\right)P^{-1}\\ & = P\mathrm{diag}\left(\sum_{k = 0}^K a_k\lambda_1^k,\ldots,\sum_{k = 0}^K a_k\lambda_n^k\right)P^{-1}\\ & \underset{K \rightarrow +\infty}{\longrightarrow} P\mathrm{diag}(f(\lambda_1),\ldots,f(\lambda_n))P, \end{align*} and this limit exists because each $\lambda_i$ verifies $|\lambda_i| \leqslant \rho < R$.
Now, in the general case, write $X = Y + N$ using Jordan-Chevalley decomposition ($Y$ is diagonalisable with the same eigenvalues as $X$ and $N$ is nilpotent, $Y$ and $N$ commute). $N$ is nilpotent thus $N^n = 0$ and the commutativity implies that we can use Newton binom formula to get that for all $k$, $$ X^k = \sum_{i = 0}^k \binom{k}{i}N^iY^{k - i} = \sum_{i = 0}^n \binom{k}{i}N^iY^{k - i}, $$ with the convention $\binom{k}{i} = 0$ when $i > k$. Therefore, for all $K \geqslant n$, \begin{align*} \sum_{k = 0}^K a_kX^k & = \sum_{k = 0}^K a_k\sum_{i = 0}^n \binom{k}{i}N^iY^{k - i}\\ & = \sum_{i = 0}^n N^i\sum_{k = i}^K a_k\binom{k}{i}Y^{k - i}\\ & = \sum_{i = 0}^n N^i\sum_{k = 0}^{K - i} a_{k + i}\binom{k + i}{i}Y^k \end{align*} Now, let for all $i$, $f_i(x) = \sum_{k \geqslant 0} a_{k + i}\binom{k + i}{i}x^k$. $k \mapsto \binom{k + i}{i}$ is a polynomial in $k$ of degree $i$ hence it doesn't affect the radius of convergence of $f_i$, which equals $R$. By the first case ($Y$ is diagonalisable), $f_i(Y)$ exists and when $K \rightarrow +\infty$ in the previous equality, we obtain that $f(X)$ exists with, $$ f(X) = \sum_{i = 0}^n N^if_i(Y) = \sum_{i = 0}^n N^i\sum_{k \geqslant 0} a_{k + i}\binom{k + i}{i}Y^k. $$ It proves that $f$ is well defined on the open set $\{X \in \mathrm{Mat}_n(\mathbb{C})|\rho(X) < R\}$ and we probably have uniform convergence on all compact subsets, just like complex series.