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My approach: They both will not be same because if we try to find the maximum likelihood function, we will get different values for both the functions.

Is this correct?

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    What's the $\arg \max$ for either function? Or, to put it in plain words: If we only care about what yields the maximum value - do we care for the actual value, or only the argument? – 2017-02-07
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    @Roland: Having the same argmax is not enough for them to be equivalent. – 2017-02-07

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No, that is a bit misleading. Likelihood is a relative concept, and what is most relevant is not the value of the likelihood function, but the subset of the parameter space for which the likelihood is maximal (in the case of maximum likelihood estimation).

In other words, while the maximum likelihood of $\theta^2$ is $1$ in the first case and $100$ in the second, in both cases, the maximum is attained at $\theta = 1$, and it is this value that is of interest.

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    @MichaelHardy You're being pedantic. I specifically said "in the case of maximum likelihood estimation." I did not at any time suggest that maximum-likelihood estimation is the only thing that is done with likelihoods. I kept my the scope of my answer limited and specific. I challenge you to find any factually incorrect statement in my post. I didn't say that the two likelihood functions were the same. – 2017-02-07
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    I would be factually incorrect to say that the reason you give is the standard reason for considering two likelihoods equivalent. – 2017-02-08
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$$ \frac{\theta^2}{100\theta^2} = \frac 1 {100}, \text{ and this does not depend on } \theta. $$

Two likelihood functions are in many situations considered equivalent, or "the same", if one of them is a constant multiple of the other. "Constant" means not depending on $\theta.$

This means that not only will they yield the same maximum-likelihood estimators, but both will yield the same posterior distributions when given the same prior distributions.

For example, suppose $\Theta$ is uniformly distributed on $[0,1]$, so that $f_\Theta(\theta) = \begin{cases} 1 & \text{if } 0\le \theta\le 1, \\ 0 & \text{otherwise.} \end{cases}$

Then let $X$ be the number of successes in $8$ independent trials with probability $\Theta$ of success on each trial, and let $Y$ be the number of independent trials needed to get $5$ successes, again with probability $\Theta$ of success on each trial. If we observe $X=5,$ the likelihood is $$ L(\theta) = \binom 8 5 \theta^5(1-\theta)^3 56\theta^5(1-\theta)^3. $$ If we observe $Y=8$, then the likelihood is $$ L(\theta) = \binom 7 4 \theta^4(1-\theta)^3\cdot\theta = 35\theta^5(1-\theta)^3. $$ In either case the posterior density is $$ f_{\Theta\,\mid\,X=5} (\theta) = 504 \theta^5(1-\theta)^3 = f_{\Theta\,\mid\,Y=8} (\theta), $$ where the constant $504$ comes from the fact that $$ \int_0^1 \theta^5(1-\theta)^3\,d\theta = \frac 1 {504}. $$