$\newcommand{\x}{\mathbf x}$$\newcommand{\one}{\mathbf 1}$$\newcommand{\X}{\mathbf X}$@kjetil b halvorsen's linked answer explains what's happening, but here's an algebraic answer just for the case of ridge regression (since there's a closed form for the solution).
Suppose we have $X\in\mathbb R^{n\times (p+k)}$ as our feature matrix where
$$
X = (\underbrace{\x \mid \x \mid \dots \mid \x}_{k\text{ times}} \mid Z)
$$
with $Z\in\mathbb R^{n\times p}$ containing whatever other columns we may have. Note that $X\beta = XP^TP\beta$ where $P$ is a permutation matrix so this shows how it's safe to assume a particular ordering of the columns of $X$ so long as we correspondingly permute the elements of $\beta$. I'll use $\X = (\x\mid \dots\mid \x) \in \mathbb R^{n\times k}$.
We have
$$
\hat\beta_R = (X^TX + \lambda I_{p+k})^{-1}X^Ty \\
= \left[\begin{array}{c|c}
\x^T\x \one_k\one_k^T + \lambda I_k & \X^TZ \\ \hline
Z^T\X & Z^TZ + \lambda I_p
\end{array}\right]^{-1}
\left[\begin{array}{c}\X^Ty \\ \hline Z^Ty\end{array}\right].
$$
I'll invert that matrix as a 2x2 block matrix (and we know it's always invertible). Letting
$$
\left[\begin{array}{c|c}
\x^T\x \one_k\one_k^T + \lambda I_k & \X^TZ \\ \hline
Z^T\X & Z^TZ + \lambda I_p
\end{array}\right] = \begin{bmatrix} A & B \\ C & D\end{bmatrix}
$$
we have
$$
\left[\begin{array}{cc} A & B \\ C & D\end{array}\right]^{-1} = \left[\begin{array}{c|c}
(A-BD^{-1}C)^{-1} & -(A-BD^{-1}C)^{-1}BD^{-1} \\\hline \cdot & \cdot\end{array}\right]
$$
where I've marked the lower row of blocks with $\cdot$ since we don't need those for the $k$ coefficients of $\x$. Letting $H_Z = Z(Z^TZ + \lambda I)^{-1}Z^T$ be the hat matrix for a ridge regression just on $Z$, we can show that
$$
BD^{-1}C = \X^TH_Z \X = \x^T H_Z \x \one\one^T
$$
and
$$
A = \x^T\x\one\one^T + \lambda I_k
$$
so
$$
A - BD^{-1}C = \x^T(I-H_Z)\x\one\one^T + \lambda I_k.
$$
Then
$$
BD^{-1} = \X^T Z(Z^TZ + \lambda I_p)^{-1}
$$
so all together we get
$$
\hat\beta_{R;k} = \left(\x^T(I-H_Z)\x\one\one^T + \lambda I_k\right)^{-1}\x^T (I-H_Z) y \one.
$$
Factoring out the scalar $\x^T(I-H_Z)\x$ this can be written as
$$
\hat\beta_{R;k} = \frac{\x^T(I-H_Z)y}{\x^T(I-H_Z)\x}\left(\one\one^T + \nu I_k\right)^{-1}\one
$$
where for convenience I'm letting
$$
\nu = \frac{\lambda}{\x^T(I-H_Z)\x}.
$$
This inverse can be done explicitly via Sherman-Morrison so
$$
\left(\one\one^T + \nu I\right)^{-1}\one = \nu^{-1}(I - (\nu + k)^{-1}\one\one^T)\one\\
= \frac{1}{\nu + k}\one.
$$
This means
$$
\hat\beta_{R;k} = (\nu + k)^{-1}\frac{\x^T(I-H_Z)y}{\x^T(I-H_Z)\x}\one \\
= \frac{\x^T(I-H_Z)y}{k\x^T(I-H_Z)\x + \lambda}\one
$$
where I've substituted back in what $\nu$ is equal to. This shows that each coefficient of $\x$ is the same thing, and that it is a shrunken form of what we'd get if $\x$ wasn't copied (if there was just one $\x$, like in the standard setting, the $k$ would disappear but the rest would be the same).
This also shows what the sum of these coefficients is:
$$
\hat\beta_{R;k} ^T\one = \frac{\x^T(I-H_Z)y}{k\x^T(I-H_Z)\x + \lambda}\one^T\one \\
= \frac{\x^T(I-H_Z)y}{\x^T(I-H_Z)\x + \lambda/k}
$$
so the coefficients don't quite sum to what we'd have if $\x$ wasn't copied, but rather their sum is what we'd get if we just had one $\x$ but divided the regularization parameter by $k$.
