Proceeding along the same lines as hexomino, I get a different result. The expectation
is undefined for two reasons. First of all, you only have an expectation if you have a probability distribution, and to get a probability distribution in this scenario you need to start with a "prior" probability distribution in advance of seeing the outcome. Second, if we use the "obvious" improper uniform prior (i.e., assume all N equally likely) then our "posterior" probability distribution is also improper (i.e., you can't make the "probabilities" add up to a finite value).
The place where I (ha!) diverge from hexomino's answer is
where he rewrites $\frac{N(N-1)\cdots(N-9)}{N^{10}}$ as $\frac{(N-1)!10}{(N-8)!N^{10}}$, where I think he's made a sign error -- it should be $\frac{(N-1)!10}{(N-10)!N^{10}}$. This means that the $N$th value is, for large $N$, roughly proportional to $1/N$, whose sum diverges.
We can try to deal with this inconvenient situation in three ways. First,
we can give up on computing an expectation and compute something else that does make sense for an improper probability distribution. For instance, the mode ("maximum likelihood estimate"). This turns out to be at $N=51$, though the value at $N=52$ is very, very close to being the same.
Second,
we can adopt a different "general-purpose" prior. For instance, we might use the "logarithmic prior" where the prior "probability" of $N$ is taken to be $1/N$. (For many cases where a variable is known to take positive values only, this is a reasonable approach, though the usual handwavy arguments for it don't really apply here.) In this case, the posterior probability for $N$ is roughly proportional to $1/N^2$, whose sum is finite, so we do get an actual probability distribution -- but, alas, not one that has an expectation, because the sum we'd need to compute for that diverges again. Still, in this case we can stil compute the mode (as above) and also the median. The mode is at $N=29$ and the median is anywhere between $N=77$ and $N=78$. (The tail of this distribution is rather "fat", which is why the median is so much bigger than the mode. The mean, again, is infinite -- it's more sensitive than the median to that fat tail.)
Third,
we can make use of our actual knowledge about cat picture websites and try to come up with a more realistic prior based on that knowledge. There are an awful lot of fairly plausible ways to do this, and unfortunately they will all give different answers. The upside is that if our prior falls off rapidly enough for large $N$ (which it will -- indeed, since presumably only finitely many cat pictures have ever been taken, it is zero for large enough $N$) then everything converges and we can compute expectations to our hearts' content. For instance, suppose we choose a prior that's uniform on $[1,M]$ and zero for $N>M$; then there's a rather ugly explicit-ish formula that for large $M$ is approximately $M/\log M$, but that limit is approached very slowly -- e.g., for $M=10^5$ the expectation is about 14000 while the approximation is about 8700. For any of these distributions with $M\ge51$ the mode ("maximum a posteriori estimate") will still be 51, because this probability distribution is just a rescaled truncation of the improper distribution we got at the outset. Or suppose we (rather arbitrarily) make the prior distribution for $N$ a Poisson distribution of mean $\mu$; then it turns out that for any plausible choice of $\mu$ the expectation comes out rather close to $\mu$ -- there's more information in the prior than in the likelihoods, so to speak.
The upshot of all this is
that I really don't think there's such a thing as a correct answer to the question. As it stands the answer is undefined for two different reasons; we can fix up both of them by specifying a suitable prior, but it turns out that different (but still fairly reasonable) priors can give extremely different answers; and typically the expectation, median and mode come out quite different, indicating that no single figure is going to serve very well to describe our actual beliefs after seeing what we saw.
Credit where due: hexomino did a lot of this before I did (with, unfortunately, an error, but we all make mistakes); all the actual calculations underlying the discussion above were done for me by Mathematica.
Incidentally,
in the real world my estimate would probably not be any of the above. I would expect a website offering cat pictures to have rather a lot of them, since the internet is made of cats, and therefore I'd guess that the repetition was most likely the result of some quirk in their software that made the images not be truly selected at random. (E.g., maybe they seed a random number generator with something involving the time, in some broken way that means that two requests exactly a minute apart are likely to produce the same output. Or something.) If I actually cared, I would take a lot more samples before trying to do any calculations.