This question appeared in a past exam paper, in the form:
Let $X = (X_1\dotsc X_n)\in\mathbb{R}^n$ be an i.i.d. sample from $U[0, \theta], \theta>0$
Apply Rao-Blackwell's theorem to the unbiased estimator $2X_1$ using the statistic $X_{(n)} = \max\{X_i\}$ to compute the efficient estimator for $\theta$. (In other parts of the problem we showed that $X_{(n)}$ is a complete sufficient statistic)
My working looks like this:
\begin{align} F_{X_{(n)}}(y) &= P(X_{(n)}\leq y)= F_{X_1}(y)^n \\ &= \left(\frac{y}{\theta}\right)^n\mathbf{1}(0\leq y\leq \theta) + \mathbf{1}(y>\theta)\\ f_{X_{(n)}}(y) &= \frac{ny^{n-1}}{\theta^n}\mathbf{1}(0\leq y\leq \theta)\\ F_{(X_{1},X_{(n)})}(x, y) &= P(X_1\leq x, X_{(n)}\leq y)\\ &= \left\{ \begin{array}{cl} \theta^{-n}xy^{n-1} & :0\leq x\leq y\leq \theta \\ \theta^{-n}y^n &: 0\leq y\leq x\leq \theta \end{array} \right.\\ f{(X_{1},X_{(n)})}(x, y) &= \theta^{-n}(n-1)y^{n-2}\mathbf{1}(0\leq x\leq y\leq \theta)\\ f_{X_{1}|X_{(n)}}(x| y) &=\frac{\theta^{-n}(n-1)y^{n-2}}{\theta^{-n}ny^{n-1}}\mathbf{1}(0\leq x\leq y)\\ &= \frac{n-1}{n}y^{-1}\mathbf{1}(0\leq x\leq y)\\ \mathbf{E}(X_1|X_{(n)}=y) &=\frac{n-1}{n}\int_0^y\frac{x}{y}dy \\ &=\frac{n-1}{n}\frac{y}{2} \\ \mathbf{E}(2X_1|X_{(n)}) &= \frac{n-1}{n} X_{(n)} \end{align} According to Rao-Blackwell's theorem , this should yield an unbiased estimator for $\theta$. Unfortunately, it is not unbiased and also yields an impossible value of $\theta$, since $X_{(n)}<\theta$.
Playing around with the result, I found that $\frac{n+1}{n} X_{(n)}$ does the job and makes sense, but I can't figure out where I went wrong in my calculations and would be very grateful if someone could point my error out.
The mistake you made is slightly subtle, especially if one is used to consider (wrongly) that distributions are either continuous or discrete but never a mix of both... Let me try to explain what is going on.
To simplify notations, assume that $\theta=1$ and call $Z=(X_1,X_{(n)})$. Your computation of $F_Z$ is correct but then you consider that the distribution of $Z$ has a density $f_Z$ while this is only partly true in the sense that a first part of the distribution of $Z$, of mass $\frac{n-1}n$, indeed has the density you write down as $f_Z$, but that there exists a second part, with no density, corresponding to the event $[X_1=X_{(n)}]$, of mass $\frac1n$.
The most convenient way to describe the distribution of $Z$ might be to specify that, for every measurable bounded function $u$, $$ E(u(Z))=\iint_{\mathbb R^2} u(x,y)f(x,y)\,\mathrm dx\mathrm dy+\int_{\mathbb R} u(x,x)g(x)\,\mathrm dx, $$ with $$ f(x,y)=(n-1)y^{n-2}\,\mathbf 1_{0\lt x\lt y\lt 1},\qquad g(x)=x^{n-1}\,\mathbf 1_{0\lt x\lt1}. $$ Note that $f$ and $g$ are nonnegative and such that $$ \iint_{\mathbb R^2} f(x,y)\,\mathrm dx\mathrm dy=\frac{n-1}n,\qquad\int_\mathbb R g(x)\,\mathrm dx=\frac1n, $$ whose sum is $1$, hence the identity above, valid for every suitable $u$, indeed defines a distribution.
The distribution of $X_{(n)}$ has, as you computed, density $h$, where $$ h(y)=\int_0^yf(x,y)\,\mathrm dx+g(y)=ny^{n-1}\,\mathbf 1_{0\lt y\lt1}. $$ In this context, a formulation of the conditional distribution of $X_1$ conditionally on $[X_{(n)}=y]$, for $y$ in $(0,1)$, is $$ P(X_1\in\mathrm dx\mid X_{(n)}=y)=\mu_y(\mathrm dx), $$ where $$ \mu_y(\mathrm dx)=\frac{f(x,y)}{h(y)}\,\mathrm dx+\frac{g(x)}{h(y)}\,\delta_y(\mathrm dx)=\frac{n-1}n\frac1y\,\mathbf 1_{0\lt x\lt y}\,\mathrm dx+\frac1n\,\delta_y(\mathrm dx). $$ This means, in particular, that $$ E(X_1\mid X_{(n)})=v(X_{(n)}), $$ where, for $y$ in $(0,1)$, $$ v(y)=\int_\mathbb Rx\,\mu_y(\mathrm dx)=\int_0^yx\,\frac{n-1}n\frac1y\,\mathrm dx+\frac1n\,y=\frac{n+1}{2n}\,y, $$ hence $$ E(X_1\mid X_{(n)})=\frac{n+1}{2n}\,X_{(n)}, $$ which reconciles your result with Rao-Blackwell theorem. We happy.