Solution:
The required numbers are $376$ and $625$.
The square of $n$ has the same last three digits as $n$ if and only if $n^{2}-n=n(n-1)$ is divisible by $1000=2^{3} \cdot 5^{3}$. Since $n$ and $n-1$ are coprime, only one of the two is even and only one of the two can be divisible by $5$. Therefore we have the following possibilities:
- $n$ is divisible both by $2^{3}$ and by $5^{3}$, hence $n$ is divisible by $1000$: no three-digit number $n$ has this property;
- $n-1$ is divisible both by $2^{3}$ and by $5^{3}$, hence $n-1$ is divisible by $1000$: no three-digit number $n$ has this property;
- $n$ is divisible by $2^{3}$ and $n-1$ is divisible by $5^{3}$; the only three-digit number with these properties is $n=376$, (a direct check indeed gives $n^{2}=141376$);
- $n$ is divisible by $5^{3}$ and $n-1$ is divisible by $2^{3}$; the only three-digit number with these properties is $n=625$, (a direct check indeed gives $n^{2}=390625$).

Solution:
Let us represent $n$ as $ABC$, where $A, B, C$ are its decimal digits. We must obtain $ABC^{2}=XYZABC$. Observe that, in general, the last digit of $n^{2}$ depends only on the last digit of $n$, just as the last two digits of $n^{2}$ depend only on the last two digits of $n$.
By direct verification, the units digit $C$ can only be $0,1,5,6$.
If $C=0$, then $n$ is divisible by $10$, so $n^{2}$ is divisible by $100$ and hence also $B=0$. But then $n$ is divisible by $100$, $n^{2}$ is divisible by $10000$ and hence also $A=0$, which is not allowed.
If $C=1$, since $(10B+1)^{2}=100B^{2}+20B+1$, we must have $2B=B$, that is $B=0$, or $2B=10+B$, which is impossible. If $B=0$, since $(100A+1)^{2}=10000A^{2}+200A+1$, we obtain as above $A=0$, that is $n=1$, which is not allowed.
If $C=5$, since $(10B+5)^{2}=100B^{2}+100B+25$, we must have $B=2$. Similarly, from $(100A+25)^{2}=10000A^{2}+5000A+625$ we obtain $A=6$. Indeed, $625^{2}=390625$ and hence $n=625$ is a solution.
If $C=6$, since $(10B+6)^{2}=100B^{2}+120B+36$, we must have $2B+3=B$, which is impossible, or $2B+3=10+B$, that is $B=7$. Finally, from $(100A+76)^{2}=10000A^{2}+15200A+5776$, we obtain $2A+7=A$, which is impossible, or $2A+7=10+A$, that is $A=3$. Indeed, $376^{2}=141376$ and hence $n=376$ is a solution.