In Wikipedia they write that the well-known OU process
$$\tag{1}
x_t=x_0\,e^{-\theta t}+\mu(1-e^{-\theta t})+\sigma\int_0^te^{-\theta(t-s)}\,dW_s
$$
can be written as
$$\tag{2}
x_t=x_0\,e^{-\theta t}+\mu(1-e^{-\theta t})+\frac{\sigma}{\sqrt{2\theta}}\,W_{1-e^{-2\theta t}}\,.
$$
(a) $\quad$ When the same Brownian motion is used in (1) and (2) the resulting OU processes $x_t$ constructed by these equations do not agree pathwise.
It is easy to see that their expectation and variance agree. Therefore, since they are Gaussian, their distribution agrees.
(b) $\quad$ To put this differently, if we want the two OU processes to agree pathwise then the BMs in (1) and (2) cannot be the same. This was nicely shown by user6247850 in a comment: The Brownian term in (1) depends on the entire path $[0,t]\ni s\mapsto W_s$ whilst the Brownian term in (2) depends only on the path $[0,1]\ni s\mapsto W_s$ (take $s=1-e^{-2\theta t}$ which is in $[0,1]$ for all $t$).
(c) $\quad$ There is an infinitude of scalings $h(t)$ and time changes $T_t$ such that, with the same BM as in (1), the OU process
$$\tag{3}
x_t=x_0\,e^{-\theta t}+\mu(1-e^{-\theta t})+\sigma\,h(t)\,W_{T_t}
$$
has the same distribution as in (1).
(d) $\quad$ Using the quadratic variation (5) below there is a standard way (e.g. [1]) of time changing another BM $B_t$ such that the OU process
$$\tag{4}
x_t=x_0\,e^{-\theta t}+\mu(1-e^{-\theta t})+\sigma\,e^{-\theta t}B_{\langle M\rangle_t}
$$
agrees pathwise with (1).
To see (a) simply compare the variances:
From (1) or (2) it is easy to see that $x_t$ has variance
$$
{\rm Var}[x_t]=\sigma^2\int_0^te^{-2\theta(t-s)}\,ds=\sigma^2e^{-2\theta t}\frac{e^{2\theta t}-1}{2\theta}=\sigma^2\frac{1-e^{-2\theta t}}{2\theta}\,.
$$
To see (c) observe that the variance of (3) is
$$
{\rm Var}[x_t]=\sigma^2\,h^2(t)\,T_t\,.
$$
In (2) we had $T_t=1-e^{-2\theta t}$ and $h^2(t)=1/(2\theta)$.
In your question you had $T_t=t$ and $h^2(t)=(1-e^{-2\theta t})/(2\theta t)\,.$
These are obviously not the only possibilities. There is an infinitude of function pairs $h(t)$ and $T_t$
that yield the correct variance. The pairs only have to obey the equation
$$
h^2(t)\,T_t=\frac{1-e^{-2\theta t}}{2\theta}\,.
$$
To see (d) we follow the standard steps for time-changed Brownian motion [1]. With the BM from (1)
$$
M_t=\int_0^te^{\theta s}\,dW_s
$$
is a martingale with quadratic variation
$$\tag{5}
\langle M\rangle_t=\int_0^te^{2\theta s}\,ds=\frac{e^{2\theta t}-1}{2\theta}\,.
$$
Writing
$$
T_t=\frac{\ln(1+2\theta t)}{2\theta}
$$
for the inverse function of $\langle M\rangle_t$ it turns out that
$$
B_t=M_{T_t}
$$
is a martingale with quadratic variation $\langle M\rangle_{T_t}=t\,,$ in other words, $B_t$ is a Brownian motion. To put it differently,
$$
M_t=B_{\langle M\rangle_t}
$$
is a time-changed Brownian motion. This gives (4)
[1] I. Karatzas, S. Shreve, Brownian Motion and Stochastic Calculus.
