I refer to this answer for background information on the $\chi^2$ test for contingency tables Chi-Square-Test: Why is the chi-squared test a one-tailed test?.
The test statistic is $X^2 = \sum_{i,j}\frac{(o_{ij}-e_{ij})^2}{e_{ij}}$,
in your case the $o_{ij}$ are the elements of your matrices $d1,d2,d3$. $e_{ij}=p_{i*} p_{*j} N$ where $p_{i*}$ and $p_{*j}$ are the (estimates of) the marginal probabeilities and $N=\sum_{i,j} o_{ij}$.
If we multiply $o_{ij}$ by a constant factor $k$ then we get $o_{ij}^{(n)}=k \cdot o_{ij}$. It is easy to show that $e_{ij}^{(n)}=k \cdot e_{ij}$, therefore
${X^{(n)}}^2 = \sum_{i,j}\frac{(o_{ij}^{(n)}-e_{ij}^{(n)})^2}{e_{ij}^{(n)}}=\frac{k^2}{k} X^2=k\cdot X^2$.
In other words after you multiply your matrix $d1$ with a constant, the test statistic $X^2$ is multiplied by the same constant ${X^{(n)}}^2 =kX^2$. This is what you see in R when you use the option correct=FALSE in chisq.test (no continuity correction, because the above formula is for the $\chi^2$ test without continuity correction):
# for d1 X2=267.8571
chisq.test(d1, correct=FALSE)
# for d1/50 X2=267.8571 /50
d2 = d1/50
d2
chisq.test(d2, correct=FALSE)
So the test statistic ${X^{(n)}}^2$ changes (is multiplied by $k$) , but the degrees of freedom of the test $(r-1)(c-1)=2$ (r is the number of rows, c the number of columns) has remained the same and therefore the p-value changes.
Is that normal ? Well what you do when you multiply the table with a factor $k$ is 'simulating' a change in sample size, and as the sample size increases, the results will become more significant:
#significant at 1pct level
chisq.test(d1, correct=FALSE)
#smaller sample size, insignificant at 1pct level
d<-d1/50
chisq.test(d, correct=FALSE)
# even worse, some cells have expected counts below 5 if sample too small and then problems with $\chi^$ approximation
d<-d1/100
chisq.test(d, correct=FALSE)
By the above, the answer to your second question is "yes".
For the first question I point to the hypothesis that is tested in the $\chi^2$ test for contingency tables: $H_0: \text{ the row-variable is independent of the column variable }$ versus $H_1: \text{ the row-variable is dependent of the column variable }$.
I refer to the answer What follows if we fail to reject the null hypothesis?, to illustrate that 'statitistical evidence' can only be found for $H_1$, so to answer your first question, with the $\chi^2$ test you can find statitistical evidence that the row variable and the column variable are dependent (when you reject $H_0$) however, if you can not reject $H_0$ then you have to conclude that 'there is no evidence that they are dependent so we accept the assumption that they are independent'. So you can not 'detect' that there is 'no association'.
In your comments you say that $H_0$ is rejected (for $d1$) although there was no difference between type 2 and type 3. However, $\chi^2$ tests whether the row variable and the column variable are dependent. 'Type 2' and 'Type 3' are values that can be taken by the column variable.
So with the $\chi^2$ test based on your table $d1$ you conclude that the 'yes/no' answer depends on the variable 'type'. To find differences for the different values of the variable 'type' you will have to analyse the terms of $X^2$ namely $\frac{(o_{ij}-e_{ij})^2}{e_{ij}}$ or if you want the sign of 'deviation' then $\frac{(o_{ij}-e_{ij})}{\sqrt{e_{ij}}}$.
If you are sure that there is no difference between 'type 2' and 'type 3' then you can, or you should, work with a $2 \times 2$ table where the last two columns of $d1$ are aggregated.
