Suppose $Y=T-nX_{(1)}$ where $T=\sum\limits_{i=1}^n X_i$.
Then a complete sufficient statistic for $\theta=(\mu,\sigma)$ based on the sample $X_1,\ldots,X_n$ is $U=(X_{(1)},Y)$. The UMVUE of $P(X_1>t)$ for a fixed $t>\mu$ is given by $$E_{\theta}[I_{X_1>t}\mid U]=P_{\theta}(X_1>t\mid U)$$
Clearly $P_{\theta}(X_1>t\mid U)=1$ for $X_{(1)}>t$.
For $X_{(1)}\le t$, noting that $U$ and $(X_1-X_{(1)})/Y$ are independent by Basu's theorem,
\begin{align}
P_{\theta}(X_1>t\mid U=u)&=P\left(\frac{X_1-X_{(1)}}{Y}>\frac{t-x_{(1)}}{y}\,\Big|\, U=u\right)
\\&=P\left(\frac{X_1-X_{(1)}}{Y}>\frac{t-x_{(1)}}{y}\right)
\\&=\sum_{j=1}^n P\left(\frac{X_1-X_{(1)}}{Y}>\frac{t-x_{(1)}}{y}\,\Big|\,X_j=X_{(1)}\right)P(X_j=X_{(1)})\tag{1}
\end{align}
Now $P(X_j=X_{(1)})=\frac1n$ for every $j=1,\ldots,n$ as the $X_i$'s are i.i.d continuous.
But $P\left(\frac{X_1-X_{(1)}}{Y}>\frac{t-x_{(1)}}{y}\,\Big|\,X_1=X_{(1)}\right)=0$ and the conditional probabilities given $X_j=X_{(1)}$ are all equal for $j=2,\ldots,n$. So $(1)$ reduces to
\begin{align}
P_{\theta}(X_1>t\mid U=u)&=\frac{n-1}{n} P\left(\frac{X_1-X_{(1)}}{Y}>\frac{t-x_{(1)}}{y}\,\Big|\,X_n=X_{(1)}\right)
\\&=\left(1-\frac1n\right) P\left(\frac{X_1-X_{(1)}}{\sum_{i=1}^{n-1}(X_i-X_{(1)})}>\frac{t-x_{(1)}}{y}\,\Big|\,X_n=X_{(1)}\right)
\end{align}
Now consider that given $X_n=X_{(1)}$, the variables $X_i-X_{(1)}$ are i.i.d exponential with mean $\sigma$ for $i=1,\ldots,n-1$ (see related post) which would imply that $(X_1-X_{(1)})/\sum_{i=1}^{n-1}(X_i-X_{(1)})$ conditioned on $X_n=X_{(1)}$ has a certain Beta distribution.
