One can do slightly better than the strategy proposed by Mike Earnest and obtain a $\frac{143}{300}\approx 48\%$ chance of success, which is $\frac{1}{300}$ more likely to succeed than the linked strategy. The strategy is as follows:
Let $f(N)=\max(1,N-1)$. Van Helsing should look to see if that are any two coffins $c_1$ and $c_2$ such that lockets $N_1$ and $N_2$ inside each respectively have $f(N_1)=c_2$ and $f(N_2)=c_1$. If so, he should switch $N_1$ and $N_2$. Otherwise, if there is any coffin $c$ with a locket $N$ inside such that $f(N)\neq c$, but such that there is some locket $N'$ with $f(N')=c$, he should switch $N'$ and $N$. He should do nothing if neither of those conditions are met.
After hearing the number $N$ which Dracula announces, Jonathan will open the coffin labelled $f(N)$.
Examples: If Van Helsing looked through the coffins numbered $1$ to $5$ and found lockets $1,\,3,\,5,\,4,\,2$ in that order, he would switch lockets $5$ and $4$ to get $1,\,3,\,4,\,5,\,2$, since now locket $4$ is in coffin $f(4)=3$ and locket $5$ is is coffin $f(5)=4$. In fact, every locket except $2$ is in the appropriate coffin, so Jonathan will find the proper locket with probability $4/5$. If the lockets were $1,\,2,\,3,\,4,\,5$, then Van Helsing would switch lockets $2$ and $3$ (or various other possibilities), since it is not possible to switch a pair of lockets into their rightful place, but it is possible to switch locket $3$ into the coffin $2=f(3)$. This leaves the arrangement $1,\,3,\,2,\,4,\,5$, where only two lockets are in the appropriate place, so Jonathan succeeds with probability $2/5$ - exactly when $1$ or $3$ is called. The last possibility is that Van Helsing does nothing, which would happen in an arrangement like $2,\,3,\,4,\,5,\,1$, where each coffin contains an appropriate locket. Note that locket $1$ is still out of place, but switching it with locket $2$ would move locket $2$ out of place, so makes no improvement. Jonathan again succeeds with probability $4/5$ here.
This succeeds with probability $\frac{143}{300}$. We will show this and that it is optimal below the horizontal line.
In particular, let us work through this game backwards. We can see that Jonathan only knows one number $N$, so his strategy is characterized by the function $f(N)$ which tells him which coffin to open given $N$.
Van Helsing's goal is therefore to maximize the number of $N$ such that the locket numbered as $N$ is in fact in the coffin labelled $f(N)$. His optimal strategy can easily be seen to be to act as follows, where he looks at only the first case below whose conditions are satisfied:
If there exist two lockets $N_1$ and $N_2$ such that $N_1$ is in coffin $f(N_2)$ and $N_2$ is in coffin $f(N_1)$, he should switch lockets $N_1$ and $N_2$, improving the probability of success by $2/5$.
Otherwise, if there exists a coffin $c$ containing a locket $N$ such that $f(N)\neq c$, but there is some $N'$ with $f(N')=c$, he should switch lockets $N$ and $N'$, improving the probability of success by $1/5$.
Otherwise, every coffin $c$ in the image of $f$ contains a locket $N$ with $f(N)=c$. In this case, the probability cannot be improved, so Van Helsing should do nothing.
Let us say that the first case happens with probability $P_1$, the second with probability $P_2$ and the third with probability $P_3$. Note the the probability of success if Van Helsing did nothing is $\frac{1}5$ since then the locket in the chosen coffin would be distributed uniformly randomly. Adding this baseline to the expected improvement in the probability due to Van Helsing gives the probability of success as
$$\frac{1}5 + \frac{1}5P_2 + \frac{2}5P_1.$$
Note that, as $P_2$ is not natural to calculate, we may use the relation $P_1+P_2+P_3=1$ and various simplifications to rewrite the probability of success as
$$\frac{2-P_3+P_1}{5}$$
To calculate $P_1$, let us define the function $S(c)=|f^{-1}[\{c\}]|$. That is, $S(c)$ is the number of lockets $N$ which Jonathan will search coffin $c$ for. This is the only aspect of $f$ that matters.
Let us first calculate the probability of a given pair of coffins $c_1$ and $c_2$ containing lockets $N_1$ and $N_2$ such that swapping the lockets puts both in their proper place. That is, we want to know the probability that $f(N_1)=c_2$ and $f(N_2)=c_1$. This probability may be seen to be $S(c_1)S(c_2)/20$.
As we are interested in the probability of this being the case for any coffins, it is useful to consider the probability that both the pairs of coffins $(c_1,c_2)$ and $(c_3,c_4)$ could be switched to the same advantage. This probability will be $S(c_1)S(c_2)S(c_3)S(c_4)/120$, since we are determining that four locations must contain elements of disjoint groups. Usefully, since a set of four coffins $\{c_1,c_2,c_3,c_4\}$ may be partitioned into two groups of two in $3$ ways, the sum of the probabilities of there being two good switches among these four coffins is $S(c_1)S(c_2)S(c_3)S(c_4)/40$.
Obviously, having only $5$ coffins, one cannot have three possible pairs of good switches, as pairs of switches may not overlap. Thus, using the inclusion-exclusion principle, we calculate $P_1$ as follows
$$P_1=\frac{\sum\limits_{\{c_1,c_2\}}S(c_1)S(c_2)}{20}-\frac{\sum\limits_{\{c_1,c_2,c_3,c_4\}}S(c_1)S(c_2)S(c_3)S(c_4)}{40}$$
where the sums run over the subsets of the coffins in the image of $f$ of the desired size.
We may calculate $P_3$ in a similar manner. In particular, if one enumerates the coffins $c_i$ in the image of $f$ as $c_1,c_2,\ldots,c_n$, then the probability of each coffin $c_i$ containing an $N_i$ with $f(N_i)=c_i$ will be $\frac{S(c_1)S(c_2)\ldots S(c_n)}{5\cdot 4 \cdot \ldots \cdot (5-n+1) }$.
If $f$ is a bijection, then $S(c)=1$ for every $c$, so we calculate that $P_1=\frac{{5\choose 2}}{20}-\frac{{5\choose 4}}{40}=\frac{15}{40}$ and $P_3=\frac{1}{5!}$ so that the overall probability of success is $\frac{71}{150}$. This choice corresponds exactly to Mike Earnest's strategy, where essentially $f(n)=n$ was used.
However, we get a better value if we have $f$ have an image of size $4$ - then, Jonathan will only ever look in four coffins, where $S$ takes the values of $1,\,1,\,1,\,2$. We then get $P_3=\frac{2}{5!}$ and $P_1=\frac{9}{20}-\frac{2}{40}=\frac{2}5$ yielding a rate of success of $\frac{143}{300}$.
Lacking a clever argument to rule out $f$ with smaller images, note that when $f$ has an image of $3$ coffins and $S$ takes the values of $1,\,1,\,3$ or $1,\,2,\,2$, then $P_1$ is the same as in the case of $f$ having an image of $4$, but $P_3$ is larger, decreasing the rate of success. If the image of $S$ is no more than two coffins, it's easy to see that $P_1$ will be at most $\frac{3}{10}$, which necessarily gives a lower probability of success than the other strategies. Thus, we settle on $f$ having an image of size $4$ to get the best success rate.
